DAMAGE BOOK
Pages are missing
00
166046
Higher Mathematics
for
Engineers and Physicists
BY
IVAN S. SOKOLNIKOFF, Ph.D.
Professor of Mathematics, University of Wisconsin
AND
ELIZABETH 8. SOKOLNIKOFF, Ph.D.
Formerly Instructor in Mathematics, University of Wisconsin
SECOND EDITION
FOURTH IMPRESSION
McGRAW-HILL BOOK COMPANY, INC.
NEW YORK AND LONDON
1941
HIGHER MATHEMATICS FOB ENGINEERS AND PHYSICISTS
COPYRIGHT, 1934, 1941, BY TUB
McGRAw-HiLL BOOK COMPANY, INC.
PRINTED IN THE UNITED STATES OF AMERICA
All rights reserved. This book, or
parts thereof, may not be reproduced
in any form without permission of
the publishers.
THE MAPLE PRESS COMPANY, YORK, PA
PREFACE
The favorable reception of the First Edition of this volume
appears to have sustained the authors' belief in the need of a
book on mathematics beyond the calculus, written from the
point of view of the student of applied science. The chief
purpose of the book is to help to bridge the gap which separates
many engineers from mathematics by giving them a bird's-eye
view of those mathematical topics which are indispensable in
the study of the physical sciences.
It has been a common complaint of engineers and physicists
that Ae usual courses in advanced calculus and differential
equations place insufficient emphasis on the art of formulating
physical problems in mathematical terms. There may also be a
measure of truth in the criticism that many students with pro-
nounced utilitarian leanings are obliged to depend on books
that are more distinguished for rigor than for robust uses of
mathematics.
This book is an outgrowth of a course of lectures offered by
one of the authors to students having a working knowledge of the
elementary calculus. The keynote of the course is the practical
utility of mathematics, and considerable effort has been made to
select those topics which arc of most frequent and immediate use
in applied sciences and which can be given in a course of one
hundred lectures. The illustrative material has been chosen for
its value in emphasizing the underlying principles rather than
for its direct application to specific problems that may confront
a practicing engineer.
In preparing the revision the authors have been greatly aided
by the reactions and suggestions of the users of this book in both
academic and engineering circles. A considerable portion
of the material contained in the First Edition has been rear-
ranged and supplemented by further illustrative examples, proofs,
and problems. The number of problems has been more than
doubled. It was decided to omit the discussion of improper
integrals and to absorb the chapter on Elliptic Integrals into
vi PREFACE
much enlarged chapters on Infinite Series and Differential
Equations. A new chapter on Complex Variable incorporates
some of the material that was formerly contained in the chapter
on Conformal Representation. The original plan of making
each chapter as nearly as possible an independent unit, in order
to provide some flexibility and to enhance the availability of the
book for reference purposes, has been retained.
I. S. S.
E. S. S.
MADISON, WISCONSIN,
September, 1941.
CONTENTS
PAGE
PREFACE v
CHAPTER I
SECTION INFINITE SERIES
1. Fundamental Concepts 1
2. Series of Constants . ..... 6
3. Series of Positive Terms . 9
4. Alternating Series ' 15
5. Series of Positive and Negative Terms . . . 16
6. Algebra of Series . ... 21
7. Continuity of Functions Uniform Convergence . 23
8. Properties of Uniformly Convergent Series . . .28
9. Power Series . 30
10. Properties of Power Series 33
11. Expansion of Functions in Power Series ... 35
12. Application of Taylor's Formula . ... 41
J3. Evaluation of Definite Integrals by Means of Power Series ... 43
14. Rectification of Ellipse. Elliptic Integrals .47
15. Discussion of Elliptic Integrals . . . 48
16. Approximate Formulas in Applied Mathematics . . 55
CHAPTER II
FOURIER SERIES
17. Preliminary Remarks . . . . 63
18. Dinchlet Conditions. Derivation of Fourier Coefficients .... 65
19 Expansion of Functions in Fourier Series . 67
20 Sine and Cosine Series . .... 73
21. Extension of Interval of Expansion 76
22. Complex Form of Fourier Series . 78
23. Differentiation and Integration of Fourier Series 80
24. Orthogonal Functions 81
CHAPTER III
SOLUTION OF EQUATIONS
25. Graphical Solutions 83
26. Algebraic Solution of Cubic 86
27. Some Algebraic Theorems 92
28. Homer's Method 95
riii CONTENTS
JBCTION PAGE
29. Newton's Method . . . 97
30. Determinants of the Second and Third Order 102
31. Determinants of the nth Order. . ... . 106
32. Properties of Determinants . . . 107
33. Minors . . . . . 110
34. Matrices and Linear Dependence 114
35. Consistent and Inconsistent Systems of Equations 117
CHAPTER IV
PARTIAL DIFFERENTIATION
36. Functions of Several Variables . 123
37. Partial Derivatives . 125
38. Total Differential i27
39. Total Derivatives . 130
40. Euler's Formula 136
41. Differentiation of Implicit Functions . 137
42. Directional Derivatives 143
43. Tangent Plane and Normal Line to a Surface 146
44. Space Curves 149
45. Directional Derivatives in Space 151
46. Higher Partial Derivatives 153
47. Taylor's Series for Functions of Two Variables 155
48. Maxima and Minima of Functions of One Variable 158
49. Maxima and Minima of Functions of Several Variables . 160
50. Constrained Maxima and Minima . .163
51. Differentiation under the Integral Sign 167
CHAPTER V
MULTIPLE INTEGRALS
'52. Definition and Evaluation of the Double Integral 173
53. Geometric Interpretation of the Double Integral 177
54. Triple Integrals ... 179
55. Jacobians. Change of Variable 183
56. Spherical and Cylindrical Coordinates 185
57. Surface Integrals . . 188
58. Green's Theorem in Space . . . 191
59. Symmetrical Form of Green's Theorem . . . 194
CHAPTER VI
LINE INTEGRAL
60. Definition of Line Integral . . 197
61. Area of a Closed Curve 199
62. Green's Theorem for the Plane . . . 202
63. Properties of Line Integrals 206
64. Multiply Connected Regions . . . . . 212
65. Line Integrals in Space . 215
66. Illustrations of the Application of the Line Integrals . .217
CONTENTS ix
SECTION PAGE
CHAPTER VII
ORDINARY DIFFERENTIAL EQUATIONS
67. Preliminary Remarks . ... 225
68. Remarks on Solutions . ... 227
69. Newtonian Laws . . 231
70. Simple Harmonic Motion . 233
71. Simple Pendulum . . 234
72. Further Examples of Derivation of Differential Equations 239
73. Hyperbolic Functions 247
^4. First-order Differential Equations 256
75. Equations with Separable Variables . 257
*f6. Homogeneous Differential Equations . . 259
77. Exact Differential Equations 262
78. Integrating Factors . 265
79. Equations of the First Order in Which One of the Variables Does
Not Occur Explicitly . 267
80. Differential Equations of the Second Order 269
81. Gamma Functions . 272
82. Orthogonal Trajectories 277
'83. Singular Solutions . ... 279
84. Linear Differential Equations . 283
85. Linear Equations of the First Order . . 284
86. A Non-linear Equation Reducible to Linear Form (Bernoulli's
Equation) . . 286
87 Linear Differential Equations of the nth Order . 287
88 Some General Theorems . . ... 291
89. The Meaning of the Operator
Z>» + oiJ)»-i + : + an-iD + anf(x) ' . . 295
90. Oscillation of a Spring and Discharge of a Condenser 299
91. Viscous Damping 302
92. Forced Vibrations . . ... 308
93. Resonance . 310
94. Simultaneous Differential Equations . . 312
95. Linear Equations with Variable Coefficients . .... 315
96. Variation of Parameters . . 318
97. The Euler Equation . . 322
98. Solution in Series . ... 325
99. Existence of Power Series Solutions . 329
100. BesseTs Equation 332
101. Expansion in Series of Bessel Functions 339
102. Legendre's Equation . . . . 342
103. Numerical Solution of Differential Equations 346
CHAPTER VIII
PARTIAL DIFFERENTIAL EQUATIONS
104. Preliminary Remarks . . .... ... 350
105. Elimination of Arbitrary Functions . . . . . . 351
x CONTENTS
SECTION PAGE
106. Integration of Partial Differential Equations. . . 353
107. Linear Partial Differential Equations with Constant Coefficients . 357
108. Transverse Vibration of Elastic String . . . 361
109. Fourier Series Solution ... . 364
110. Heat Conduction . 367
111. Steady Heat Flow ..... 369
112. Variable Heat Flow ... . . . 373
113. Vibration of a Membrane . 377
114. Laplace's Equation . . ... .... 382
115. Flow of Electricity in a Cable . . 386
CHAPTER IX
VECTOR ANALYSIS
116. Scalars and Vectors . ... 392
117. Addition and Subtraction of Vectors 393
118. Decomposition of Vectors. Base Vectors . 396
119. Multiplication of Vectors . . 399
120. Relations between Scalar and Vector Products 402
121. Applications of Scalar and Vector Products . . . 404
122. Differential Operators . . . 406
123. Vector Fields . . 409
124. Divergence of a Vector . 411
125. Divergence Theorem . . . . . 415
126. Curl of a Vector . 418
127. Stokcs's Theorem . 421
128. Two Important Theorems 422
129 Physical Interpretation of Divergence and Curl 423
130. Equation of Heat Flow 425
131. Equations of Hydrodynamics 428
132. Curvilinear Coordinates . . 433
CHAPTER X
COMPLEX VARIABLE
133. Complex Numbers . ... . . 440
134. Elementary Functions of a Complex Variable . . 444
135. Properties of Functions of a Complex Variable 448
136. Integration of Complex Functions .... ... 453
137. Cauchy's Integral Theorem . ... 455
138. Extension of Cauchy's Theorem . .... 455
139. The Fundamental Theorem of Integral Calculus . 457
140. Cauchy's Integral Formula . . . . . 461
141. Taylor's Expansion. . . . ... . ... 464
142. Conformal Mapping .... . . . . 465
143. Method of Conjugate Functions . . 467
144. Problems Solvable by Conjugate Functions 470
145. Examples of Conformal Maps . ... ... 471
146 Applications of Conformal Representation . . . 479
CONTENTS xi
SECTION PAGE
CHAPTER XI
PROBABILITY
147. Fundamental Notions . .... 492
148. Independent Events . . 495
149. Mutually Exclusive Events . 497
150. Expectation. . . . 500
151. Repeated and Independent Trials . . . 501
152. Distribution Curve 504
153. Stirling's Formula . 508
154. Probability of the Most Probable Number . . .511
155. Approximations to Binomial Law . 512
156. The Error Function ... 516
157.* Precision Constant. Probable Error . .521
CHAPTER XII
EMPIRICAL FORMULAS AND CURVE FITTING
158. Graphical Method . . 525
159. Differences 527
160. Equations That Represent Special Types of Data 528
161. Constants Determined by Method of Averages 534
162. Method of Least Squares . 536
163. Method of Moments 544
164. Harmonic Analysis . . ... 545
165. Interpolation Formulas . . . 550
166. Lagrange's Interpolation Formula . 552
167. Numerical Integration 554
168. A More General Formula 558
ANSWERS . . 561
INDEX . . 575
HIGHER MATHEMATICS
FOR ENGINEERS AND
PHYSICISTS
CHAPTER I
INFINITE SERIES
It is difficult to conceive of a single mathematical topic that
occupies a more prominent place in applied mathematics than
the subject of infinite series. Students of applied sciences meet
infinite series in most of the formulas they use, and it is quite
essential' that they acquire an intelligent understanding of the
concepts underlying the subject.
The first section of this chapter is intended to bring into
sharper focus some of the basic (and hence more difficult) notions
with which the reader became acquainted in the first course in
calculus. It is followed by ten sections that are devoted to a
treatment of the algebra and calculus of series and that represent
the minimum theoretical background necessary for an intelligent
use of series. Some of the practical uses of infinite series are
indicated briefly in the remainder of the chapter and more fully
in Chaps. II, VII, and VIII.
1. Fundamental Concepts. Familiarity with the concepts
discussed in thig section is essential to an understanding of the
contents of this chapter.
FUNCTION. The variable y is said to be a function of the variable
x if to every value of x under consideration there corresponds at least
one value of y.
If x is the variable to which values are assigned at will, then
it is called the independent variable. If the values of the variable
y are determined by the assignment of values to the independent
1
2 MATHEM&TICS FOR ENGINEERS AND PHYSICISTS §1
variable x, then y is called the dependent variable. The functional
dependence of y upon x is usually denoted by the equation*
V = /(*)•
Unless a statement to the contrary is made, it will be supposed
in this book that the variable x is permitted to assume real
values only and that the corresponding values of y are also real.
In this event the function f(x) is called a real function of the real
variable x. It will be observed that
(1-1) y =
does not represent a real function of x for all real values of x, for
the values of y become imaginary if x is negative. In order that
the symbol f(x) define a real function of x, it may be necessary to
restrict the range of values that x may assume. Thus, (1-1)
defines a real function of x only if x ^ 0. On the other hand,
y — \/x2 — 1 defines a real function of x only if \x > I.
SEQUENCES AND LIMITS. Let some process of construction
yield a succession of values
Xij 2*2, #3, , XH) ,
where it is assumed that every xt is followed by other terms.
Such a succession of terms is called an infinite sequence. Exam-
ples of sequences are
(a) 1, 2, 3, • • • , n, • • • ,
,,,1 11 1 / iw i !
(6) 2' - 41 8' ~ 16' ' ' ' '<-*> V
(c) 0,2,0, 2, ••-, ! + (-!)", • • • .
Sequences will be considered here only in connection with the
theorems on infinite series, t and for this purpose it is necessary
to have a definition of the limit of a sequence.
DEFINITION. The sequence x\y x%, • • • , xn, • • • is said to
converge to the constant L as a limit if for any preassigned positive
number «, however small, one can find a positive integer p such that
\xn — I/I < e for all n > p.
* Other letters are often used. In particular, if more than one function
enters into the discussion, the functions may be denoted by /i(x), ft(x), etc.;
by/(aO, g(x), etc.; by F(x)t G(x), etc.
t For a somewhat more extensive treatment, see I. S, §okolnikoff?
Advanced Calculus, pp. 3-21 f
§1 INFINITE SERIES 5
which is convergent to the value 2. In order to establish this
fact, note that
is a geometric progression of ratio J^, so that
J_
on 7~ ^ TvHTZTi"*
Heftce, the absolute value of the difference between 2 and sn
is l/2n~1, which can bo made arbitrarily small by choosing n
sufficiently large.
On the other hand, if x — — 1, the series (1-4) becomes
which does not converge; for s2n = 0 and S2n~i = 1 for any choice
of n and, therefore, lim sn does not exist. Moreover, if x = 2,
n— » oo
the series (1-4) becomes
1 + 2 + 4 + • • • + 2-1 + • • • ,
so that sn increases indefinitely with n and lim sn does not exist.
n — » oo
If an infinite, series does not converge for a certain value of x,
it is said to diverge or be divergent for that value of x. It will
be shown later that the series (1-4) is convergent for — I < x < 1
and divergent for all other values of x.
The definition of the limit, as given above, assumes that the
value of the limit $ is known. Frequently it is possible to infer
the existence of S without actually knowing its value. The
following example will serve to illustrate this point.
Example. Consider the series
and compare the sum of its first n terms
_!,!,!, , JL
Sn - l + 2! + 3! + ' ' ' + n!
with the sum of the geometrical progression
6 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
S. = 1 + \ + ± + • • • + ~
The corresponding terms of Sn are never less than those of $„; but, no
matter how large n be taken, Sn is less than 2. Consequently, s« < 2;
and since the successive values of sn form an increasing sequence of
numbers, the sum of the first series must be greater than 1 and less than
or equal to 2. A geometrical interpretation of this statement may help
to fix the idea. If the successive values of s»,
Si = 1,
«2 = 1 + 21 = 1.5,
ss - 1 + 5j 4- ^ = 1.667,
*4=*l + || + ^ + jj = 1.708,
s* = 1 + I + |j + jj + ^ = 1.717,
are plotted as points on a straight line (Fig. 1), the points representing
the sequence Si, $2, * • • , sn, • • - always move to the right but never
0 ,, 1 15 1667 2
* - FIG. 1.
progress as far as the point 2. It is intuitively clear that there must be
some point s, either lying to the left of 2 or else coinciding with it, which
the numbers sn approach as a limit. In this case the numerical value
of the limit has not been ascertained, but its existence was established
with the aid of what is known as the fundamental principle.
Stated in precise form the principle reads as follows: // an infinite
set of numbers si, §2, * • * , sn, • • • forms an increasing sequence (that is,
SN > Sn, when N > n) and is such that every sn is less than some fixed
number M (that is, sn < M for all values of n\ then sn approaches a limit
s that is not greater than M (that is, lim sn = s < M). The formulation
n— » oo
of the principle for a decreasing sequence of numbers «i, s2, • • * ,
sn, • • • , which are always greater than a certain fixed number w, will
be left to the reader.
2. Series of Constants. The definition of the convergence of
a series of functions evidently depends on a study of the behavior
§2 INFINITE SERIES 7
of series of constants. The reader has had some acquaintance
with such series in his earlier study of mathematics, but it seems
desirable to provide a summary of some essential theorems that
will be needed later in this chapter. The following important
theorem gives the necessary and sufficient condition for the
convergence of an infinite series of constants:
00
THEOREM. The infinite series of constants £ un converges if
n = l
and only if there exists a positive integer n such that for all positive
integral values of p
\Sn+p — Sn\ SS \Un+l + Un+2 + ' ' ' + Un+f>\ < €,
where e is any preassigned positive constant.
The necessity of the condition can be proved immediately by recalling
the definition of convergence. Thus, assume that the series converges,
and let its sum be Sy so that
lim sn = S
n— > oo
and also, for any fixed value of p,
lim sniP = S.
n— > oo
Hence,
lim (sn+p — sn) = lim (un+} + un+z + • • • + un+p) = 0,
n— * « n— * *
which is another way of saying that
for a sufficiently large value of n.
The proof of the sufficiency of the condition requires a fair degree
of mathematical maturity and will not be given here.*
This theorem is of great theoretical importance in a variety of
investigations, but it is seldom used in any practical problem
requiring the testing of a given series. A number of tests for
convergence, applicable to special types of series, will be given in
the following sections.
It may be remarked that a sufficient condition that a series
diverge is that the terms un do not approach zero as a limit when
n increases indefinitely. Thus the necessary condition for con-
vergence of a series is that lim un = 0, but this condition is not
n— * «o
* See SOKOLNIKOFF, I. S., Advanced Calculus, pp. 11-13.
8 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §2
sufficient; that is, there are series for which lim un = 0 but which
n—> «
are not convergent. A classical example illustrating this case
is the harmonic series
in which Sn increases without limit as n increases.
Despite the fact that a proof of the divergence of the harmonic
series is given in every good course in elementary calculus, it will
be recalled here because of its importance in subsequent con-
siderations. Since
_1_ + _J_+ . . • ! ^ ! !
i "i I l r> l
1 n + n " " 2n 2'
it is possible, beginning with any term of the series, to add a
definite number of terms and obtain a sum greater than V£.
If n = 2,
3 + 4 > 2;
n =' 4,
n = 8,
1, , J_ , , J_ 1.
9 "*" 10 + ' ' ' + 16 > 2;
n = 16,
_1_J_ .±^1
17 + 18 + ' ' ' + 32 > 2
Thus it is possible to group the terms of the harmonic series
in such a way that the sum of the terms in each parenthesis
exceeds ££,; and, since the series
1+2: + ! + 2:+ '••
is obviously divergent, the harmonic series is divergent also.
§8 INFINITE SERIES 9
3. Series of Positive Terms. This section is concerned with
series of the type
an = a! + az + ' ' ' an • • * ,
1
where the an are positive constants. It is evident from the
definition of convergence and from the fundamental principle
(see Sec. 1) that the convergence of a series of positive constants
will be established if it is possible to demonstrate that the partial
sums sn remain bounded. This means that there exists some
positive number M such that sn < M for all values of n.
The proof of the following important test is based on such a
demonstration.
oo
COMPARISON TEST. Let 2 an be a series of positive terms,
n = l
00
and let 2 bn be a series of positive terms that is known to converge.
n = l
00
Then the series 2 an is convergent if there exists an integer p such
n = l
00
that, for n > p, an ^ bn. On the other hand, if 2 cn is a series of
n-l
positive terms that is known to be divergent and if an ^ cn for
oo
n > p, then 2 #n is divergent also.
n = l
Since the convergence or , divergence of a series evidently is not
affected by the addition or subtraction of a finite number of terms, the
proof will be given on the assumption that p = 1. Let sn = «i + a2
00
+ • - • + an, and let B denote the sum of the series 2 bn and Bn its
n = l
nth partial sum. Then, since a» ^ bn for all values of n, it follows that
sn ^ Bn for all values of n. Hence, the sn remain bounded, and the
oo
series 2 dn is convergent. On the other hand, if an ^ cn for all values of
n=»l
00 00
n and if the series 2 cn diverges, then the series 2 an will diverge also.
n=l n=l
There are two series that are frequently used as series for
comparison.
a. The geometric series
(3-1) a + ar + ar2 + • • • + ar» + • • • ,
10 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §3
which the reader will recall* is convergent to . _ if \r\ < 1
and is divergent if \r\ > 1.
6. The p series
(3-2) l+g+£+ •••+£+•••,
which converges if p > 1 and diverges if p < 1.
Consider first the case when p > 1, and write (3-2) in the form
(3-3)
4-
T
(2n _
where the nth term of (3-3) contains 2n~1 terms of the series
(3-2). Each term, after the first, of (3-3) is less than the corre-
sponding term of the series
1 J_ O . Jl __ L A . ___ __ L . . . 4_ On— 1
1 ^ ^ ^ ^ ^
2? ^ 4p (2n~1)p '
or
(3-4) 1 + Tj^Zi + /2P-1)2 + ' ' ' + /2p-l)n-l + ' ' ' *
Since the geometric series (3-4) has a ratio l/2p~l (which is less
than unity for p > 1), it is convergent and, by the comparison
test, (3-2) will converge also.
If p = 1, (3-2) becomes the harmonic series which has been
shown to be divergent.
If p < 1, l/np > l/n for n > 1, so that each term of (3-2),
after the first, is greater than the corresponding term of the
harmonic series; hence, the series (3-2) is divergent also.
Example 1. Test the series
1 + I + I+ ... +!+-..
A T 22 T 33 T -r nn -r
The geometric series
1,1,1, _• 1 .
1 + 22 + 23 + * ' ' + 2* + ' " '
* Since the sum of the geometric progression of n terms a -f or -f or2
i i -i • , . a — arn a X1 N
4- • • • H- arn r is equal to ^_ — = __ (1 — rn).
§3 INFINITE SERIES 11
is known to be convergent, and the terms of the geometric series are
never less than the corresponding terms of the given series. Hence, the
given series is convergent.
Example 2. Test the series
iog~2 loi~3 log! ' k^ •
Compare the terms of this series with the terms of the p series for
given series is divergent, for its terms (after the first) are greater
than the corresponding terms of the p series, which diverges when
p = 1.
00
RATIO TEST. The series 2 «n of positive terms is convergent if
,. U'n-f-l ^ +
lim — — = r < 1
n— * °° an
and divergent if
lim ^^ > 1.
n— » QO dn
// lim -^ = 1, the series may converge or diverge.
Consider first the case when r < 1, and let q denote some constant
between r and 1. Then there will be some positive integer N such that
— < q for all n £ N.
Hence.
aNq,
and
Since q < 1, the series in the right-hand member is convergent'; there-
fore, the series in the left-hand member converges, also. It follows that
00
the series S ctn is convergent*.
n-l
-
If the limit of the ratio is greater than 1, then an+i > an for every
00
n ^ N so that lim an j& 0, and hence the series S an is divergent.
n— » «o n = 1
12 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §3
It is important to observe that this theorem makes no reference
to the magnitude of the ratio of an+\/an but deals solely with tLr
limit of the ratio. Thus, in the case of the harmonic series the
ratio is an+\/an = n/(n + 1), which remains less than 1 for all
finite values of n, but the limit of the ratio is precisely equal
to 1. Hence the test gives no information in this case.
Example 1. For the series
and, therefore, the series converges.
Example 2. The series
1+1L + 1L+ . . . +"j*L+ .
10 102 103 10"
is divergent, for t. •
r an+i ,. (n + l)!10rt , n
lim -^i- = lim v , ' ' -- r- = lim
-- , -- r- — ~r
an n-^oo 10n+1 n! n-K» 10
Example 3. Test the series
Here
r fln+1
"
5-6
T^ T 7
1
2n - l)2n '
. (2n - l)2n
4n2
- l)(2n + 2)
- 2n
lim ^ X "" ^
4n' +
6n + 2 ,
^"l +1-4
2n
1
"2n2
Hence, the test fails; but if the given series be .compared with the
p series for p = 2, it is seen to be convergent* J*-
oo *
CAUCHY'S INTEGRAL TEST. Let 2 an be a series of positive
n = l
terms such that an+i < an. If there exists a positive decreasing
function f(x), for x > 1, such that f(n) = an, then the given series
converges if the integral
exists; the series diverges if the integral does not exist.
INFINITE SERIES
13
The proof of this test is deduced easily from the following
graphical considerations. Each term an of the series may be
thought of as representing the area of a rectangle of base unity
and height /(n) (see Fig. 2). The sum of the areas of the first
n inscribed rectangles is less than f"+1f(x) dx, so that
f(x) dx.
But f(x) is positive, and hence
l f(x) dx
f(x) dx.
If the integral on the right exists, it follows that the partial sums
are bounded and, therefore, the series converges (see Sec. 1).
T»vC
The sum of the areas of the circumscribed rectangles, a\ + a 2
+ • • • + an, is greater than f "+1 f(x) dx] hence, the series will
diverge if the integral does not exist.
Example 1. Test the harmonic series
In this case, f(x) — - and
x
f °° 1 Cn dx
I - dx = lim I — = lim log n
Jl X n-» oo Jl X n-> oo
and the series is divergent.
14 MATHEMATICS FOR EN&NEERS AND PHYSICISTS §3
00 i
Example 2. Apply Cauchy's test to the p series ]5£ — - where p > 0.
^^ p
Taking f(x) = — i observe that
-p
l xf l_p- ,. "p*1-
= log a;|?, if p = 1.
/« ^XJ.
— exists if p > 1 and does not exist if p < 1.
or*
PROBLEMS
1. Test for convergence
l
— __
2 2 • 22 3 • 23 4 • 24
-
/ ^ i i 2! , 3! ,
(c) 1 + 2i + 35 +
1 ,2,3,
^ + v + 2"3 + • • • ;
21^2 3l^g~3 4Togl "
2. Use Cauchy's integral test to investigate the convergence of
' --
v1'/' A ~ i _j_ 22 ~ 2 + 32
00
3. Show that the series 2 an of positive terms is divergent if nan
has a limit L which is different from 0. Hint: Let lim nan = L so that
nan > L — c for n large enough. Hence, an >
n— » oo
- —
§4 INFINITE SERIES 15
4. Test for convergence
1
(2n + I)2'
n
4. Alternating Series. A series whose terms are alternately
positive and negative is called an alternating series. There is a
simple test, due to Leibnitz, that establishes the convergence of
many of these series.
TEST FOR AN ALTERNATING SERIES. // the alternating series
a\ — c&2 + «a — a* + * * • , where the at are positive, is such
that an+i < an and lim an = 0, then the series is convergent.
n— > oo
Moreover, if S is the sum of the series, the numerical value of the
difference between S and the nth partial sum is less than an+i.
Since
S2n = (ai - a2) + (as — flu) + • • • + («2n-i - a2n)
= ai — (a2 — a3) — • • • — (a2n-2 — a2n-i) — «2n,
it is evident that $2n is positive and also that $2n < #1 for all
values of n. Also, $2 < $4 < *e < * * • , so that these partial
16 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §6
sums tend to a limit S (by the fundamental principle). Since
«2n+i = «2n + «2n+i and lim a2n+i = 0, it follows that the partial
n— » oo
sums of odd order tend to this same limit. Therefore, the series
converges. The proof of the second statement of the test will
be left as an exercise for the reader.
Example 1. The series
is convergent since lim - = 0 and — — - < -• Moreover. 54
n-> * n n + 1 n
— % + H — y± differs from the sum S by less than y*>.
Example 2. The series
--..-.-...
2 "*" 31 4 "*" 32 6 "*" 33
is divergent. Why?
6. Series of Positive and Negative Terms. The alternating
series and the series of positive constants are special types of the
general series of constants in which the terms can be either posi-
tive or negative.
DEFINITION. // u\ + u<t + • • • + un + • • • is an infinite
series of terms such that the series of the absolute values of its terms,
\Ui\ + \u%\ + • • ' + \un\ + • • • , is convergent, then the series
Ui -f- u<i + • • • + un + • • • is said to be absolutely convergent.
If the series of absolute values is not convergent, but the given series
is convergent, then the given series is said to be conditionally
convergent.
Thus,
l-i+i_!+i ____
2+3 4+5
is convergent, but the series of absolute values,
is not, so that the original series is conditionally convergent.
If a series is absolutely convergent, it can be shown that the
series formed by changing the signs of any of the terms is also a
convergent series. This is an immediate result of the following
theorem:
§6 INFINITE SERIES 17
THEOREM. // the series of absolute values 2 \un\ is convergent,
00
then the series 2 un is necessarily convergent.
Let
and
If pn denotes the sum of the positive terms occurring in sn and — qn
denotes the sum of the negative terms, then
(5-1) Sn = Pn - qn
and
tn = Pn + qn>
00
The series 2 \un\ is assumed to be convergent, so that
n = l
(5-2) lim tn = lim (pn -f qn) ss L.
n—» oo n— » QO
But pn and gn are positive and increasing with n and, since (5-2) shows
that both remain less than L, it follows from the fundamental principle
that both the pn and qn sequences converge. If
lim pn — P and lim qn = Q,
n — > * H — > oo
then (5-1) gives
« lim srl = lim (pn — qn) = P — Q,
00
which establishes the convergence of 2 un.
Moreover, it can be shown that changing the order of the
terms in an absolutely convergent series gives a series which is
convergent to the same value as the original series. * However,
conditionally convergent series do not possess this property. In
fact, by suitably rearranging the order of the terms of a condi-
tionally convergent series, the resulting series can be made to
converge to any desired value. For example, it is knownf that
the sum of the series
111 f_n»-i
i—i-i-l — ±j_
1 0<0 A 1
-.--
23 4 n
* See SOKOLNIKOFF, I. S., Advanced Calculus, pp. 240-241.
t See Example 1, Sec. 13.
18 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §5
p
is log« 2. The fact that the sum of this series is less than 1 and
greater than % can be made evident by writing the series as
which shows that the value of sn > % for n > 2; whereas, by
writing it as
-»• * O Of \ A f
it is clear that sn < I for n > 2. Some questions might < be
raised concerning the legitimacy of introducing parentheses in a
convergent infinite series. The fact that the associative law
holds unrestrictedly for convergent infinite series can be estab-
lished easily directly from the definition of the sum of the infinite
series. It will be shown* next that it is possible to rearrange
the series
so as to obtain a new series whose sum is equal to 1. The positive
terms of this series in their original order are
3' 5' 7' 9'
The negative terms are
11
'
g
In order to form a series that converges to 1, first pick out, in
order, as many positive terms as are needed to make their sum
equal to or just greater than 1, then pick out just enough negative
terms so that the sum of all terms so far chosen will be just less
than 1, then more positive terms until the sum is just greater
than 1, etc. Thus, the partial sums will be
*2 - - 2 = 2'
- 1,1,1 31
S4==1-2 + 3 + 5 = 30'
* General proof can be constructed along the lines of this example.
§8 INFINITE SERIES 19
l * 1 1 47
. .
2 3 5-3 7 9- 1260
1,1,11, 1,1_1_ 1093
1 2 + 3 + 5 4 + 7 "*" 9 6 ~ 1260'
It is clear that the series formed by this method will have a sum
equal to 1.
As another example, consider the conditionally convergent
series
(5-3) l--4= + -^--4-+---.
' V2 -s/3 VI
Let the order of the terms in (5-3) be rearranged to give the
series
LL 4. -1 L\
The nth term of (5-4) is
1
which is greater than
j == •*• _|_ *__ — [i — . \ •*• ,
oo
But the series S &n is divergent, and it follows that the series
n = l
(5-4) must diverge.
00
Inasmuch as the series S |wn| is a series of positive terms, the
n»l
tests that were developed in Sec. 3 can be applied in establishing
00
the absolute convergence of the series 2 un. In particular, the
n = l
ratio test can be restated in the following form:
20 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §5
00
RATIO TEST. The series S un is absolutely convergent if
and is divergent if
lim
lim
Un+l
Un
< 1
> 1.
// the limit is unity, the test gives no information.
Example 1. In the case of the series
. z2 . xs . x*
lim
xn (n — 1)!
ft! xn~l
0
for all values of x. Hence, the series is convergent for all values of
x and, in particular, the series
22 23
1 - 2 + 21 - 3J + ' ' '
is absolutely convergent.
Example 2. Consider the series
1
Here
lim
n— » oo
1 - x ^ 2(1 - xY ^ 3(1 -
. n(l - ap*
(n + 1)(1 - x)n+l 1
(n + 1)(1 - x)
I
Therefore, the series will converge if
n-i^r < 1 or 1< |1 1 x\,
\i — x\
which is true for x < 0 and for x > 2.
For x = 0 and for x — 2 the limit is unity, but if x = 0 the series
becomes the divergent harmonic series
and if x = 2 there results the convergent alternating series
§6 INFINITE SERIES . 21
— i4.i^i4... •-!-( — nn-i-u...
1 ^ 2 3 ^ ^ ' n^
It follows that the original series converges for x < 0 and for x ^ 2
and diverges for 0 < x < 2.
6. Algebra of Series. The following important theorems are
stated without proof:*
THEOREM 1. Any two convergent series
U = Ui + U2 + ' ' • + Un + ' ' '
V = Vi + V% + ' ' • + Vn + ' * '
can be added or subtracted term by term to give
U + V = (Ui + Vi) + (Ma + »2) + ' • • + (tin + »»)+••'
or
U - V = (m - v,) + (M* - f>2) + • • • + (u» - t>0 + ' ' ' •
// /Ae original series are both absolutely convergent, then the resulting
series will be absolutely convergent also.
THEOREM 2. //
U = ui + uz + • • • + un + • • •
F = fli + tf2 + ' • ' + Vn + ' • •
are two absolutely convergent series, then they can be multiplied like
finite sums and the product series will converge to UV. Moreover,
the product series will be absolutely convergent. Thus,
UV = U\V\ + UiV% + U%Vi + U\V$ + U^Vz + UzV\ + ' ' ' .
THEOREM 3. In an absolutely convergent series the positive
terms by themselves form a convergent series and also the negative
terms by themselves form a convergent series. If in a convergent
series the positive terms form a divergent series, then the series of
negative terms is also divergent and the original series is conditionally
convergent.
THEOREM 4. // u\ + u% + • • • + un + • • • is an abso-
lutely convergent series and if Mi, M^9 • • • , Mn, • • • is any
sequence of quantities whose numerical values are all less than some
positive number N, then the series
UiMl + UiMz + ' ' ' + UnMn + • ' *
is absolutely convergent.
* See SOKOLNIKOFF, I. S., Advanced Calculus, pp. 212-213, 241-245.
22 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §6
Exarppk. Consider the series
sin x __ sin 2x sin 3x __
1« 23 + 33 " ' ' ' '
This series is absolutely convergent for all values of x> for the series
p - 2~3 + 3~3 ""
is absolutely convergent and (sin nx\ ^ 1.
PROBLEMS
1. Show that the following series are divergent:
, . 5 7,9 11 , 2n + 3 ,
(«)2-4 + 6""8 + ""+ (~1} ~2^~ + ' ' ' J
(6) 2-22 + 3- 3*2 + 4-4$+ • • • ;
(c) 1 - 1 + 1 " i + ' ' ' •
2. Test for convergence, and if the series is convergent determine
whether it is absolutely convergent.
— - 1.1 1 .....
1-3-5-7 .
_ - . _
3 3-6 3-6-9 3-6-9-12'''
,,2 3,4 5
(c) i - ^ + 3 - 4 + ' ' ' •
3. For what values of x are the following series convergent?
/v«2 /y.3 /yn
W*-l+l ---- +<-D-'^+---;
/^2 /p4 /p6
(6) 1 - 2! + j| - g-, + • • • ;
(c) 1 - « + *2 - «» + • • • ;
w; + i + 3i+---+i+----
4. Determine the intervals of convergence of the following series:
iT4 2 V^Tl 3
% (6) * + 2b2 + 3lx3 + 4b4 + • • • ;
, m(m - 1) m(m - 3)(m — 2)
(c) 1 + ws H -- 21 - * "! -- 31 - x +
where w is not a positive integer.
INFINITE SERIES
23
7. Continuity of Functions. Uniform Convergence. Before
proceeding with a discussion of infinite series of functions, it is
necessary to have a clear understanding of the concept of con-
tinuity of functions. The reader will recall that a function
f(x) is said to be continuous at a point x = XQ if lim f(x) = f(xQ)
y— »xo
regardless of how x approaches XQ. From the discussion of
the limit in Sec. 1, it appears that this concept can be defined in
the following way :
DEFINITION. The function f(x) is continuous at the point
x = XQ if, corresponding to any ^reassigned positive number c,
it is possible to find a positive number 5 such that
(7-1) I/O) — /Cr0)| < « whenever \x - rr0| < 5.
The foregoing analytical definition of continuity is merely a
formulation in exact mathematical language of the intuitive
y=f(x0)-e
-y-f(x)
FIG. 3
concept of continuity. If the function f(x) is represented by a
graph and if it is continuous at the point x = #o, then it is
possible to find a strip bounded by the two parallel lines x = x0
+ 5 and x = x0 — 5, such that the graph of the function will lie
entirely within the strip bounded by the parallel lines y = /(a?o)
+ e and y = f(xQ) — e (Fig. 3). But if the function is discon-
tinuous at some point (such as x = x\), then no interval about
such a point can be found such that the graph of the function will
lie entirely within the strip of width 2e, where e is arbitrarily
small.
DEFINITION. A function is said to be continuous in an interval
(a, b) if it is continuous at each point of the interval
24 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §7
If a finite number of functions that are all continuous in an
interval (a, 6) are added together, the sum also will be a continuous
function in (a, fe). The question arises as to whether this
property will be retained in the case of an infinite series of con-
tinuous functions. Moreover, it is frequently desirable to
obtain the derivative (or integral) of a function f(x) by means of
term-by-term differentiation (or integration) of an infinite series
that defines /(x). Unfortunately, such operations are not always
valid, and many important investigations have led to erroneous
results solely because of the improper handling of infinite series.
A discussion of such questions requires an introduction of the
property of uniform convergence of a series.
It was stated in Sec. 1 that the series
(7-2) ui(x) + u*(x) + • • • + un(x) + • • •
is convergent to the value $, when x = XQ> provided that
(7-3) lim sn(xo) = S.
I n—> oo
The statement embodied in (7-3) means that for any preassigned
positive number €, however small, one can find a positive number
N such that
|«n(&o) - S\ < e for all n ^ N.
If the series (7-2) is convergent for every value of x in the interval
(a, 6), then the series (7-2) defines a function S(x). Let x0 be
some value of x in (a, b), so that
\sn(xo) — S(xo)\ < e whenever n ;> N.
It is important to note that, in general, the magnitude of N
depends not only on the choice of e, but also on the value of XQ.
This last remark may be clarified by considering the series
(7-4) x + (x - l)x + (x - l)z2 + • • •
+ (x - I)*"-1 + • • • .
Since
s»(x) = x + (x - l)x + (x - l)x2 + • • • + (x - l)^-1
= xn,
it is evident that
lim sn(x) = lim xn = 0, if 0 ^ x < 1.
§7 INFINITE $ERIES 25
Thus, S(x) = 0 for all values of x in the interval 0 <> x < 1, and
therefore
MX) - S(x)\ = |*» - 0| = \x»\.
Hence, the requirement that \sn(x) — S(x)\ < c, for an arbitrary
e, will be satisfied only if xn < e. This inequality leads to the
condition
n log x < log e.
Since log x is negative for x between 0 and 1, it follows that it is
necessary to have
logo:
which clearly shows the dependence of N on both e and x. In fact,
if e = 0.01 and x = 0.1, n must be greater than log 0.01/log 0.1
= — 2/ — 1 = 2, so that N can be chosen as any number greater
than 2. If e = 0.01 and x = 0.5, N must be chosen larger than
log 0.01/log 0.5, which is greater than 6. Since the values of
log x approach zero as x approaches unity, it appears that the
ratio log e/log x will increase indefinitely and that it will be
impossible to find a single value of 'TWwhich will serve for e = 0.01
and for all values of x in 0 ^ x < 1.
It should be noted that the discussion applies to the interval
(0, 1) and that it might be possible to find an N, depending on e
only, if some other interval were chosen. If the series and the
interval are such that it is possible to find an N, for any pre-
assigned e, which will serve for all values of x in the interval, then
the series is said to converge uniformly in the interval.
00
DEFINITION OF UNIFORM CONVERGENCE. The series S un(x)
n«l
is uniformly convergent in the interval (a, b) if, for any € > 0, there
exists a positive number N, independent of the value of x in (a, b),
such that
\S(x) - sn(x)\ < e for all n> N.
The distinction between uniform convergence and the type of
convergence exemplified by the discussion of the series (7-4)
will become apparent in the discussion of the series
(7-5) 1 + x + x* + • - • + x* + • • • ,
where — J^ < x ^ J^.
26 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §7
J — %n
Since sn(x) = -z - , it follows that
i — x
8(x) = lira «„(*) = lim T--
n— * oo n— » oo \1 •k
Then,
\8(X) - «„(*)! =
1 - x
which will be less than an arbitrary e > 0 if
\xn\ < e(l - x).
Hence,
n log |x| < log e(l — a;),
or
(7-6) n >
.
log |z|
Again, it appears that the choice of AT will depend on both
x and €, but in this case it is possible to choose an N that will
serve for all values of x in ( — J^, H)« Observing that the ratio
log e(l — a;) /log |#| assumes its maximum value, for a fixed e,
when x = — J^, it is evident that if N is chosen so that
Iog2?
then the inequality (7-6) will be satisfied for all n *t N.
Upon recalling the conditions for uniform convergence, it is
seen that the series (7-5) converges uniformly for — % < x < %.
However, it should be noted that (7-5) does not converge uni-
formly in the interval ( — 1, 1). For, in this interval, the ratio
appearing in (7-6) will increase indefinitely as x approaches the
values ±1. The discussion given above shows that the series
(7-5) is uniformly convergent in any interval ( — a, a), where
a < 1.
It may be remarked that the series (7-5) does not even con-
verge for x = ±1. For x = 1, it is obviously divergent, and
when x = — 1 the series becomes
1 - 1 + 1 - 1 +
T-5) defin
the value ^ when x = — 1.
If — 1 < x < 1, (7-5) defines the function ^ , which takes
1 "*-* X
§7 INFINITE SERIES 27
As is often the case with definitions, the definition of uniform
convergence is usually difficult to apply when the behavior of a
particular series is to be investigated. There are available
several tests for the uniform convergence of series, the simplest of
which is associated with the name of the German mathematician
Weierstrass.
THEOREM. (WEIERSTRASS M TEST). Let
(7-7) ui(x) + u2(x) + • • • + un(x) + • • •
be a series of functions of x defined in the interval (a, 6). // there
exists a convergent series of positive constants,
Mi + M* + • • • + Mn + • • • ,
such that |^t(#)| ^ Mlfor all values of x in (a, b), then the series
(7-7) is uniformly and absolutely convergent in (a, 6).
Since, by hypothesis, the series of M 's is convergent, it follows
that for any prescribed c > 0 there exists an N such that
Mn+i + Mn+i +•••<€ for all n ^ N.
But \u,(x)\ < M% for all values of x in (a, 6), so that
for all n ^ N and for all values of x in (a, b). Therefore, the
series (7-7) is uniformly and absolutely convergent in (a, b).
The fact that the Weierstrass test establishes the absolute
convergence, as well as the uniform convergence, of a series means
that it is applicable only to series which converge absolutely.
There are other tests that are not so restricted, but these tests are
more complex. It should be emphasized that a series may con-
verge uniformly but not absolutely, and vice versa.
Example 1. Consider the series
sin x sin 2x sin nx
-p- + -22- + ' ' ' + -£5- + ' ' ' •
Since |sin nx\ ^ 1 for all values of x, the convergent series
15 + Ji +•••+»» + •••
will serve as an M series. It follows that the given series is uniformly
and absolutely convergent in any interval, no matter how large.
Example 2. As noted earlier in this section, the series
1 + x + x2 + - • • + z» + • • •
28 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §8
converges uniformly in any interval (—a, a), where a < 1. The
series of positive constants
1 + a + a2 + - • • + a* + • • •
could be used as an M series in this case, since this series converges for
a < 1 and |a;l| ^ a* for x in ( — a, a).
PROBLEMS
1. Show that the series (7-4) is uniformly convergent in the interval
(0, H).
2. By using the definition of uniform convergence, show that
1 1 1
(x + l)(x 4-2) (x + n- l)(x + n)
is uniformly convergent in the interval 0 < z < 1.
ies to sh
\S(X) ~
Hint: Rewrite the series to show that sn(x) = — r- — and therefore
X ~\~ 71
" x + n
3. Test the following series for uniform convergence:
, x cos x , cos 3x , cos 5x ,
(°0 -p- + -3^" + —52- + • • • ;
.^^ sin ^cc sin QX sin u^c
(m —: 0~ H 5 Tf + "= ^~ ~T ' ' ' ',
I ' O O • O 0 * /
(c) 1 + ^ cos ^ + x1 cos 219 + xs cos 30 + • • • , \x\ ^ xl < 1 ;
. ,. cos 2x cos 3x , cos 4x
""
2" "3" 4" '
(e) lOz + 102x2 + 108OJ3 + • • • .
8. Properties of Uniformly Convergent Series. As remarked
in the preceding section, the concept of uniform convergence was
introduced in order to allow the discussion of certain properties
of infinite series. This section contains the statements* of three
important theorems concerning uniformly convergent series.
THEOREM 1. Let
be a series such that each ul(x) is a continuous function of x in
the interval (a, 6). // the series is uniformly convergent in (a, 6),
then the sum of the series is also a continuous function of x in (a, 6).
* For proofs, see I. S. Sokolnikoff, Advanced Calculus, pp. 256-262.
§8 INFINITE SERIES 29
COROLLARY. A discontinuous function cannot be represented
by a uniformly convergent series of continuous functions in the
neighborhood of the point of discontinuity.
THEOREM 2. // a series of continuous functions,
converges uniformly to S(x) in (a, 6), then *
0 S(x) dx = fftui(x) dx + f*ut(x) dx + • • • + fft un(x) dx +
ot Jet Jot Jot
where a < a < b and a < 0 < b.
THEOREM 3. Let
- • + un(x) + • • •
be a series of differentiate functions that converges to S(x) in
(a, 6) . // the series
u((x) + u'2(x) + • • • + <(s) + • • •
converges uniformly in (a, 6), then it converges to S'(x).
These theorems provide sufficient conditions only. It may be
that the sum of the series is a continuous function when the series
is not uniformly convergent. It is impossible to discuss neces-
sary conditions in this brief introduction to uniform convergence.
It may happen also that the series is differentiate or integrable
term by term when it does not converge uniformly. In the
chapter on Fourier scries it will be shown that a discontinuous
function can be represented by an infinite series of continuous
functions. In that chapter, it is established that the series
sin 2x . sin 3x
_- _g
represents the function x for —w < x < TT. But, if this series
be differentiated term by term, the resulting series is
2(cos x — cos 2x + cos 3x — • - • ),
which does not converge in (— TT, ?r); for the necessary condition
for convergence, namely, that lim \un\ = 0, does not hold for any
n— * *
value of x.
The series used in the first example of Sec. 7,
, . sin x , sin 2x sin 3x sin nx ,
(8-1) -jg I ^2 I 32 r • ' ' H ~z r • ' • ,
30 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
is uniformly convergent in any interval (a, 6) and as such defines
a continuous function S(x). Moreover, the series can be inte-
grated term by term to produce the integral of S(x). The
term-by-term derivative of (8-1) is
(8-2)
cos x + Y2 cos 2x + % cos 3x +
which is convergent in (0, TT), but the M series for (8-2) cannot
be found since I +% + }£+••• is divergent. This merely
suggests that (8-2) may not converge to the derivative of S(x),
but it does not say that it will not.
PROBLEMS
1. Test for uniform convergence the series obtained by term-by-term
differentiation of the five series given in Prob. 3 of Sec. 7.
2. Test for uniform convergence the series obtained by term-by-
term integration of the five series given in Prob. 3, Sec. 7.
9. Power Series. One of the most important types of infinite
series of functions is the power series
00
(9-1) V anxn ss
anxn
in which the at are independent of x. Some of the reasons for
the usefulness of power series will become apparent in the dis-
cussion that follows.
Whenever a series of functions is used, the first question which
arises is that of determining the values of the variable for which
the series is convergent. The ratio test was applied for this
purpose in the examples discussed in Sec. 5. In general, for a
power series,
lim
un+i
un
= lim
so that the series converges if
and diverges if
lim
n— * oo
lim
n— » ao
an-i
\0>n-l
< 1
> 1.
Therefore, the series will converge for those values of x for which
\x < lim
INFINITE SERIES
31
If lim
On-l
an
r, it follows that the series will converge when x
lies inside the interval ( — r, r), which is called the interval of con-
vergence, the number r being called the radius of convergence.
This discussion establishes the following theorem :
00
THEOREM. // the series S anxn is such that
n = 0
lim
then the series converges in the interval — r<x<r and diverges
outside this interval. The series may or may not converge at the
end points of the interval.
Example 1. Consider the series
/v.2 /*.3
/>»n
-
n
Since lim
= lim
n - 1
= 1, the series converges for
— 1 < x < 1 and diverges for \x\ > 1. At the end point x = —1 the
series becomes
.-.+S-H----
which is convergent. At the end point x = 1 the divergent series
is obtained. Hence, this power series is convergent for —1 ^ x < 1.
Example 2. The series
1 + x + 2lx2 + • • • + n\xn + - - -
will serve to demonstrate the fact that there are power series which
converge only for the value x = 0. For
lim
lim
(n - 1)!
n!
lim -
0.
Obviously, the series converges for x — 0, as does every power series,
but it diverges for every other value of x.
* Power series in x — h are frequently more useful than the
special case in which the value of h is zero. A series of this type
has the form
ai(x - h) + a2(x -
an(x -
32 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
In this case the test ratio yields
lim
= lim
\x - h\.
If this limit is less than 1, the series is convergent; if greater than
1, the series is divergent; and if the limit is equal to 1, the test
fails and the values of x, which make the limit equal to 1, must be
investigated. Thus, if the series is
then
lim
n— » oo
Un+l
= lim
n— » «
(x-
D"
(n-
I)2
un
n<
(*-
I)"-1
= lim l - - - - |s - 1| = |x - 11-
n_» w \ n n2/ ' ' '
Therefore the series converges if x — 1| < 1, orO<x<2, and
diverges for \x — 1| > 1, or x < 0, x > 2. For a; — 1 = 1, or
x = 2, the series becomes
which is the p series for p = 2 and is therefore convergent. For
==0) the series becomes
which is an alternating series of decreasing terms with lim un = 0
n— * oo
and is therefore convergent. Thus the series is convergent for
0 ^ x <. 2.
PROBLEM
Find the interval of convergence for each of the following series, and
determine its behavior at the end points of the interval:
(a) 1 + x + x2 + x3 + - • • ;
§10 INFINITE SERIES 33
(d) I - 2x + 3x* - 4z3 + • • ' ;
1
_
2 2* • 2
(* - 2) - 1 (* -
10. Properties of Power Series. The importance of power
series in applied mathematics is due to the properties given in
the theorems of this section, as will be evident from the applica-
tions discussed in succeeding sections.
THEOREM 1 . // r > 0 is the radius of convergence of a power
00
series 2 anxn, then the series converges absolutely and uniformly for
n = 0
every value of x in any interval a ^ x < b that is interior to
Since the interval (a, 6) lies entirely within the interval (— r, r), it
is possible to choose a positive number c that is less than r but greater
than a and b. The interval (a, 6) will then lie entirely within the
interval ( — c, c); and it follows that, for a ^ x ^ 6,
\anxn\ < \ancn\.
00
The series of positive constants S \ancn\ is convergent, for c < r, and,
n = 0
accordingly, can be used as a Weierstrass M series establishing the
00
absolute and uniform convergence of S anxn in (a, b).
n=0
oo
THEOREM 2. A power series 2 anxn defines a continuous func-
/ton /or aK values of x in any closed interval (a, 6) 2Aa2 is interior to
the interval of convergence of the series.
This statement is a direct consequence of the preceding theorem
and of Theorem 1, Sec. 8.
THEOREM 3. // the radius of convergence of the power series
00 _ 00
2 anxn is r, then the radii of convergence of the series 2 nanxn~1
n=0 n-O
00
and ^^ — ~-^ xn+1, obtained by term-by-term differentiation and
n = 0
integration of the given series, are also r.
34 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §10
If the radius of convergence can be determined from the ratio
test, then the proof follows immediately from the fact that if
lim
= r. then lim
/ •* \
and
nan
lim
n— * *
j
(n-
nan
Since the series obtained by term-by-term differentiation and
integration are also power series, these processes can be repeated
as many times as desired and the resulting series will be power
series that converge in the interval ( — r, r). It follows from
Theorem 1 that all these series are uniformly and absolutely
convergent in any interval which is interior to ( — r, r). How-
ever, the behavior of these series at the end points x = — r and
x = r must be investigated in each case.
For example, the series
/»»2 <i»3 /l»>i
l+x +!+!+••• •+*-+•••
has unity for its radius of convergence. The series converges
for x = —1 but is divergent for a: = 1. The series obtained by
term-by-term differentiation is
1 + x + x* + • • • + x* + - - • ,
which has the same radius of convergence but diverges at both
x = 1 and x = — 1. On the other hand, the series obtained by
term-by-term integration is
/>»2 /y.3 /y.4 /yrt-fl
+ Ji> , «*/ . U/ - . *l/ .
_ 1 _ L _ _1_ . . . I _ I ...
-^-^-^ +
which converges for both x = 1 and x = — 1.
This discussion leads to the conclusion stated in the following
theorem :
oo
THEOREM 4. A power series 2 o>nxn may be differentiated and
n«0
integrated term by term as many times as desired in any closed
interval (a, b) that lies entirely within the interval of convergence of
the given series.
00
THEOREM 5. // a power series S Q>nxn vanishes for all values
n-p
of x lying in a certain interval about the point x =• 0, then the
§11 INFINITE SERIES 35
coefficient of each power of x vanishes, that is,
do = 0, ai = 0, a2 = 0, • • • , a* = 0, • • • .
The reader may attempt to construct the proof of this theorem
with the aid of Theorem 2 of this section.
11. Expansion of Functions in Power Series. It was stated in
Theorem 2, Sec. 10, that a power series defines a continuous
function of x in any interval which lies within the interval of
convergence. This theorem suggests at once the possibility of
using such a power series for the purpose of computation. For
example, the values of sin x might be obtained by means of a
power series. Accordingly, it becomes necessary to develop
some method of obtaining such a power series, and this section
is devoted to a derivation of Taylor's formula and a discussion of
Taylor's series.
One of the simplest proofs of Taylor's formula will be given
here. * It assumes that the given function f(x) has a continuous
nih derivative throughout the interval (a, 6). Taylor's formula
is obtained by integrating this nth derivative n times in suc-
cession between the limits a andjfe where x is any point in (a, 6).
Thus,
I fM(x)dx =/<»-« (a-
)
I I /<n)(z) (dx)2 = I /<"-»(*) dx - f /<— «(o) dx
Ja Ja Ja Ja
™(x) (dx)3 =/(»-« (*) _/(— »(0) - (Z - a)/<"-2)(a)
= /(*) - /(a) -(x- a)/' (a)
2! (n - 1)! '
* For other proofs, see I. 8. Sokolnikoff, Advanced Calculus, pp. 291-295.
36 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §11
Solving for f(x) gives
(11-1) /(*) - /(a) + (x - a)/' (a) + (x ~a)>(a)
where
(11-2) Rn = f • • • f/(n)(*) (dx)*
i/a Ja
The formula given by (11-1) is known as Taylor's formula and
the particular form of Rn given in (11-2) is called the integral
form of the remainder after n terms. The foregoing can be
stated in the form of a theorem.
TAYLOR'S THEOREM. Any function f(x) that possesses a con-
tinuous derivative /(n) (x) in the interval (a, b) can be expanded in
the form (11-1) for all values of x in (a, b).
The term Rn, which represents the difference between f(x)
and the polynomial of degree n — 1 in x — a, is frequently more
useful when expressed in a different form. Since*
P/(n)(z) dx - (x* '- a)/(n)(£), where a < f < x,
Ja
repeated integration gives
(11-3) Rn = £ - - - J[* /<»>(*) (dxY = (X ^q)n/(n)(e.
The right-hand member of (11-3) is the Lagrangian form of the
remainder after n terms.
The special form of Taylor's formula that is obtained by
setting a = 0 is known as the Maclaurin formula. In this case
(11-4) f(x) =/(0)
where
* The student will recall from elementary calculus that
I <f>(x) dx == (6 — a)<f>(£), where a < £ < b.
§11 INFINITE SERIES 37
Taylor's formula with the Lagrangian form of the remainder
is often encountered in a somewhat different form, which results
from setting x — a = A. Since a < £ < x, £ can be written
in the form a + Oh, where 0 < 6 < 1. Hence, (11-1) becomes
(11-5) /(a + h) = /(a) ~
' where
In this derivation of Taylor's formula, it was assumed that
f(x) possesses a continuous nth derivative, and as a result it
appeared that then f(x) could be expressed as a polynomial of
degree n in x — a. It should be noted, however, that only the
first n coefficients of this polynomial are constants, for the
coefficient of (x — a)n is a function of £ and the value of £ is
dependent upon the choice of x. It may happen that f(x)
possesses derivatives of all orders and that the remainder Rn
approaches zero as a limit when n -— » <*> regardless of the choice
of x in (a, b). If such is the case, the infinite series
(11-6) /(a) +f'(a)(x - a) +/"(a) (x "^ + • • •
---
is convergent and, in general,* it converges to/(x).
The series given in (11-6) is called the Taylor's series expansion,
or representation, of the function f(x) about the point x = a.
The special form of (11-6) that is obtained when a = 0, namely,
(11-7)
is called Maclaurin's series.
Example. Find the Taylor's series expansion of cos x in powers of
7T
*-§•
Since
/'(*)= -sins, /'(!)= -1'*
* For a further discussion of this point, see I. S. Sokolnikoff, Advanced
Calculus, pp. 296-298.
38 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §11
/"(*) = -cos*,
/'"(*) = sin *, /
f"(x) = cos x, fiv = 0;
it follows that the result is
Since it is often possible to obtain a power series expansion
of a function f(x) by some other method, the question arises as
to the relation of such an expansion to the Taylor's series expan-
sion for f(x). For example, a power series expansion for the
function ^ - is obtained easily by division, giving
J. """* X
The reader can check the fact that the Maclaurin expansion for
this function is identical with the power series obtained by
division. That this is not an exceptional case is established in
the following theorem:
UNIQUENESS THEOREM. There is only one possible expansion
of a function in a power series in x — a; and, therefore, if such an
expansion be found in any manner whatsoever, it must coincide with
Taylor's expansion about the point a.
Suppose that f(x) could be represented by two power series in
x — a, so that
f(x) = a0 + ai(x - a) + a2(x - a)2 + • • •
+ an(x - a)n + • • •
and
f(x) = 60 + bi(x - a) + 62(z - a)2 + • • •
+ bn(x — a)n + • • • .
Since both these expansions represent f(x) in the vicinity of a,
there must be some interval about the point x = a in which
both the expansions are valid. Then, in this interval,
§11 INFINITE SERIES 39
00 00
V an(x - a)n = ]£ bn(x — a)n,
n-O n=0
or
jj (an - &»)(* - a)- = 0.
n«=0
It follows from Theorem 5, Sec. 10, that
an ~ bn - 0, (n = 0, 1, 2, • • • ),
or
an = bn> (n = 0, 1, 2, • • • ).
Hence, the two power series expansions are identical.
Taylor's formula is frequently more useful in a slightly modified
form. Let
x — a s= h,
so that
x = a + h.
Then
/(*) = /(a) +/'(a)(z - a) +^~) (x - a)2 + - • •
/<»->(a) -
+ (^nyi^ a; +
becomes
(11-8) /(a + A) = /(a) + /'(a) A + K + '
(n-l)! n! '
in which 0 < 8 < 1, so that a<a + 9h<a + h.
PROBLEMS
1. Find the expansion of each of the following functions in power
series in x:
(a) ex, (b) sin x, (c) cos x, (d) tan"1 x,
(e) sin"1 x, (/) sec #, (g) tan x, (h) e«inx.
2. Expand
(a) log a; in powers of x — 1 ;
(6) tan x in powers of a; — T;
(c) e* in powers of x — 2;
/j\ • e v
(a) sm a; m powers of x — g;
(e) 2 + x2 — 3x* + 7x* in powers of x — 1.
40 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §11
3. Show that sin x can be developed about any point a in a series
(11-8) which converges for all values of h.
4. Differentiate term by term the power series in x for sin x and thus
obtain the power series in x for cos x. What is the interval of con-
vergence of the resulting series?
x3 x5
6. Divide the series sin x = x — ^y + ^ — • • • by the series
x* x4
cos £==1— 21 "t" ZT "~ "**> anc^ ^us obtain the series for tan x.
6. Differentiate the series for sin"1 x to obtain the expansion in powers
of # for (1 — x2)~M. Find the interval of convergence. Is convergence
absolute? Investigate the behavior of the series at the end points of the
interval of convergence.
7. Establish with the aid of Maclaurin's series that
(a + &)* SE k(l + *)«• = k [l + mx + m(m2~ 1} x* + - • • ],
where m is not a positive integer.
This series is convergent for \x\ < 1 and divergent when |a;| > 1 A
complete discussion of this series will be found in Sokolnikoff's Advanced
Calculus. Some facts are.
If x — I, convergence is absolute if m > 0;
If £ = 1, convergence is conditional if 0 > m > — 1;
If x = —1, convergence is absolute if m > 0;
If x == —1, series diverges when m < 0;
If # = 1, series diverges when m ^ — 1.
oo oo
8. Let /(?/) = S bnynsmdy~ S anxn be convergent power series. If
n=0 n=0
f(y) is a polynomial, then the powers of y in terms of x can be determined
by repeated multiplications and thus the expansion for/(?/) in powers of x
can be obtained. But if f(y) is an infinite series, this procedure may not
be valid. Inasmuch as the power series in x is always convergent for x = 0
and since the value of y for x — 0 is a0, it is clear that the interval of con-
oo
vergence of S bnyn must include a0 if the series for f(y\ in powers of x,
n = 0
is to converge. But if a0 = 0, then f(y) surely can be expanded in
power series in x by this method, for the point 0 is contained in the
00
interval of convergence of S bnyn.
n=*0
Apply this method to deriving the series in powers of x for e*^ x by
setting
xs , x6
y = sm x = x - + - • • •
§12 INFINITE SERIES 41
and
Explain why this method fails to produce the series in powers of x for
log (1 + e*)i where e* = y.
12. Application of Taylor's Formula. In this section two
illustrations of the application of Taylor's formula will be given,
and in each case the remainder will be investigated to determine
the error made in using the sum of the first n terms of the expan-
sion instead of the function itself.
1. Calculate the Value of sin 10°. Since 10° is closer to 0°
than to any other value of x for which the values of sin x and its
derivatives are known, the Maclaurin expansion for sin x will
be determined and evaluated for x = 10° = IT /IS radian. Then
f(x) = /(O) + /'(O)*
n\
_
'
where 0 < £ < 75-
io
Since
f(x) = sin x, /(O) = 0;
f'(x) = oosx, /'(0) = 1;
/"(*) = - sin *, /"(O) = 0;
f"(x) = - cos x, /'"(0) = -1;
................................... y
/<»>(*) = sin (x + ~^, /<»>(0) = sin ^;
therefore,
Here,
Rn(x) s ~ /<»>($) == ~ f(n}(0x), 0 < 6
= — . sin
n!
If only the terms through x7 (or x8) are used in computing
sin 7T/18, the error will be
42 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §18
" 1 • {** ^9 \ /T\9 1 fcr
W 91 8m V Is + 2 V = Iw 9i cos ii
9!'
so that
sm 18 ~ 18 U87 3! + V18/ 5! Vl8/ 7f
with an error less than ( j^ j ^-
2. Compute the Value of e1-1. It can be established readily, by
expanding ex in Maclaurin's series and evaluating for £ = 1, that
e = 2.71828 • • • . In order to compute the value of e1-1, the
expansion of e* about x = 1 will be used. The expansion is
Since
/(*) = ex, /(I) = ^
?(*) = «-, /'(I) = e;
/<»>(*) = e-, /<•>(!) = e;
and
/f»'(0 = e«, 1 < { < x;
therefore
(x - 1)* + • • • + n J 1; (x - I)
Here
so that the error made in using only four terms is
«4 = || (X - 1)«.
If X - 1.1,
e1-1 =
= 1.105166e
§13 INFINITE SERIES 43
Thus, e1-1 = 1.105166e with an error of (0.0001/24)e*. Since £
lies between 1 and 1.1 and since e* is an increasing function, e* is
certainly less than e2. An approximate value of e2 is 7, and the
error is certainly less than 0.0007/24 = 0.00003. Therefore,
e1-1 = 1.1052e,
correct to four decimal places.
13. Evaluation of Definite Integrals by Means of Power
Series. One of the most important applications of infinite series
is their use in computing numerical values of definite integrals,
such as f J e~x* dx, in which the indefinite integral cannot be
found in closed form. Moreover, the values of many tran-
scendental functions are computed most easily by this method.
Several examples of this use of infinite series are given in this
section.
Example 1. Consider
^o+^
Since
(1 + 2)-l = 1 - Z -f 22 _
for 1 3 1 < 1, it follows that
Example 2. Since
dz
if | z | < 1, therefore
It is evident that this method of obtaining the expansion of sin"1 x is
much less complicated than the direct application of Taylor's formula.
Example 3. In order to evaluate the integral
f
Jo
2adx
- x)(2ax -
44 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §13
express it as
dx
A _ jL^"5*,
\ 2j
\
and then replace ( 1 — ^- j by its expansion in powers of ~» giving
, 1-3-5/s'
^2-4-6 \2a
If this integral is expressed as
CM / x \ dx
== +
fh I / x
Jo 2\2a
- ;2
2-4
and each integral evaluated, there results
+
••]•
This expression gives the period of the simple pendulum. By making
the change of variable x = h sin2 <p, the integral reduces to
/- 2^ J.U -*'**'*>-»**
where &2 = V2a.
This is the form used in the discussion of the simple pendulum given
in Sec. 71. In this illustration, h denotes the height of the pendulum
bob and a the length of the pendulum.
/*! ex _ e-x
Example 4. The integral J0 dx cannot be evaluated by the
usual method for evaluating a definite integral, for the indefinite integral
cannot be obtained. Moreover, the expansion for > if obtained
x
directly with the aid of Maclaurin's formula, would lead to an extremely
complicated expression for each derivative. The expansion is most
easily obtained by using the separate expansions for e* and e~*. Thus,
e..i+s + 5! + |?+...
L\ o!
§13 INFINITE SERIES 45
and
/ z3 xb
e* - e- = 2 ^ + ^ j + ^ + • •
Hence,
fl Px — e-x / 1 1
Jo — i— * - 20 + F3l + 5-1! +•••- 2'1145'
Example 5. In order to evaluate the integral J Q e*™ * dx, recall that
so that
. . sin2 a; , sin3 x
sm JP H -- 21 --- ' -- 3"!
Then
Cw ' f" ( ' . , sin2 a; , sin3 a; ,
J0 e***dx = Jo (,1 + sm » + ~2p + -3^ +••
TT
, f 5 / . sin2 x sin3 a; , . \ ,
= 2 Jo I1 + Sln * + ~2T + ~3T + ) **,
which can be evaluated with the aid of the Wallis formula
. J f^ j (^ - 1)(" - 3) • • • 2 or 1
sinn x dx — L cosn x dx — -- / - ^ - ^ - : - «,
Jo n(n — 2) • • • 2 or 1 '
where a = 1, if n is odd, and a = ^> if n is even.
In order to justify the term-by-term integration, it is sufficient to
show that the series in the integrand is uniformly convergent. That
such is the case is obvious if one considers
l + l + i\ + wi+ •••
as the Weierstrass M series.
PROBLEMS
1. Calculate cos 10°, and estimate the maximum error committed
by neglecting terms after x6.
2. Find the interval of convergence of the expansion of e* in power
series in x. Determine the number of terms necessary to compute e1-1
accurate to four decimal places from this expansion, and compare the
result with illustration 2, Sec. 12.
3. Compute sin 33°, correct to four decimal places.
46 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §13
Cx dx
4. Expand the integrand of J0 1 . gg in power series in x, and
integrate term by term. Compare the result with that of Prob. l(d),
Sec. 11. __
5. Compute v/35 = 2(1 + ?^2)^» correct to five decimal places.
6. Develop the power series in x for sin"1 x and hence establish that
_ , . - -
6 ~ 2 ~*~2 3 \2/ "" 2-4 5 \2
7. Show, by squaring and adding the power series for sin # and
cos x, that.
sin2 x + cos2 #=1.
8. Evaluate by using series expansions of the integrands
/ v f1 • / ON , ,,x TH sin a; efo
(a) JQ gin (x') dx; (6) J0 ==;
(e) §* cos (*«) dx; (/) ^ (2 - cos x)~» dx
T1
= Jo
/*l log x f0-1 log(l - z) /**
(?) Jo 9 r-r^ <*» = Jo — -z — *; « Jo
/*x
(i) I &***dx.*
9. Show, by multiplication of series, that
(1 + x + x2 + • • )2 = 1 + 2x + 3z2 +
= (1 - x)-2.
10. Expand to terms in x6
(a) \/cos a;;
sin g
(c)
cos
11. Determine the magnitude of a, if the error in the approximation
sin a === a is not to exceed 1 per cent.
a ~ sin a , . a8 a5
Hint: = 0.01 and sma = a — -^ + -=7— • • • .
a oi ol
* See form 787, B. O, Peirce, A Short Table of Integrals.
§14 INFINITE SERIES 47
14. Rectification of Ellipse. Elliptic Integrals. In spite of its
importance and apparent simplicity, the problem of finding the
length of an elliptical arc is not usually considered in elementary
calculus. This is because the integral that arises is incapable
of evaluation in terms of elementary functions. However, the
evaluation can be effected by means of series expansion of the
integrand function, as will be shown in this section.
Let the equation of the ellipse be
r2 7v2
J> + f, = l, a>b.
The length of arc from (0, 6) to (x\, y\) is given by the integral
Computing dy/dx and substituting its value in (14-1) gives
r /, _!_ b* ** A - r
~ Jo \ + rf tf^Ts<to ~ Jo
Recalling the fact that the numerical eccentricity of the ellipse
is k = -S/&2 — W/a, the integral given above can be written as
where k2 < 1.
Let x = a sin 0; then dx = a cos 6 dd, and (14-2) becomes
(14-3) 8 = a Vl - k2 sin2 6 dd.
JQ
The series expansion of the integrand function is most easily
obtained by writing it as (1 — fc2 sin2 0p and expanding by
use of the binomial theorem. Then (14-3) is replaced by
s = a J (l - ^k* sin2 0 - g fc4 sin4 6 - • • • ) d8,
and term-by-term integration* gives
* Term-by-term integration is valid here, for the series
serves as a Weierstrass M series.
48 MATHEMATICS FOR ENGINEERS AND PHYSICISTS &15
(14-4) s = a L> - ^ k* I sin2 0 dO - ^ ft4 I %in4 0d0 - • • •
L ^ J° o Jo
_ l-3.5.-.(2n-3)fc2M Psin2n 9dg _ . . . 1
2 • 4 • 6 • • • 2w Jo J
If (14-4) is used, it is possible to evaluate s for particular
values of k and <p. However, the integral in (14-3) is so impor-
tant that there are extensive tables* giving its value for* many
choices of k and <p. This integral for the value of a — 1 is
called the elliptic integral of the second kind and is denoted by
the symbol E(k, <p). If <p = 7r/2, the integral is called the com-
plete elliptic integral of the second kind, which is denoted by the
symbol E.
The elliptic integral of the second kind having been defined, it
seems desirable to mention the elliptic integral of the first kind,
although the latter arises in considering the motion of a simple
pendulum and will be discussed in more detail in Sec. 71. The
elliptic integral of the first kind, F(k, ^>), has the form
(14-5) F(k, v) = F , d°
Jo Vl - k2 sin2 S
The complete elliptic integral of the first kind, which arises
when <p = 7T/2, is denoted by the symbol K. Values of F(ky <p)
and of K are also tabulated, but the evaluation can be obtained
from (14-5) by means of series expansion of the integrand.
Thus, one has the expansion
(14-6) F(fc, ?) = <p + i fc2 I sin2 6 d6 + | fc4 | sin4 6 d6
"JO o JO
16. Discussion of Elliptic Integrals. The elliptic integral of
the first kind is a function defined by the integral
(15-1) F(k, *) m F d° . =, fc2 < L
Jo v 1 - k2 sin2 S
* See the brief table in B. O. Peirce, A Short Table of Integrals, pp.
121-123.
§16 INFINITE SERIES 49
If sin 6 is replaced by z, (15-1) becomes
(15-2) f(k, x) = (* —r— dz k* < 1.
' Jo V(l - *2)(1 - *V)
This is an alternative form of the elliptic integral of the first kind.
Similarly, the same change of variable transforms the integral
of the second kind
(15-3) E(k, *>) = I* ^/^^~k^n^e dO, fc2 < 1,
Jo
into
(15-4) i(k, x) = £ ^^f dz, k* < 1.
It will be recalled that any integral of the type
\ R(x, -\/ax2 + bx + c) dXj
where R is a rational function of the variables x and
\/ax'2 + bx + c,
is integrablb in terms of the elementary functions, i.e., power,
trigonometric, and logarithmic functions. It can be shown that
the integration of integrals of the type
(15-5) J R(x, V<™3 + bx'^~cx~+~d) dx
and
(15-6) J R(x, \/ax* + bx3 + ex2 +~dx + e) dx
requires, in general, the introduction of new functions obtained
from the elliptic integrals.
The evaluation of (15-5) and (15-6) can be reduced to the
evaluation of integrals of the elementary types and the following
new types:
a. Elliptic integral of the first kind :
fvi N Cx dz
F(k'x) m va - «•>(! - *y>- °r
de
/
VI - fc2 sin2 6
b. Elliptic integral of the second kind:
r
Jo
— 2
E(k, x) = _ dz, or
= I *
Jo
- & sin2 6 dO.
£0 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §15
c. Elliptic integral of the third kind:
dz
or
fi(n,*,*) = r
Jo
H(n, t, *) s p
Jo
de
/
(1 + n sin2 (?) \/l ~ &2 sin2
The problem of reducing the integrals of expressions involving
square roots of cubics and quartics to normal forms is not difficult,
but it is tedious and will be omitted here.* Integrals involving
y
2.0
1.0
FIG. 4.
square roots of polynomials of degree higher than the fourth lead,
in general, to more complicated functions, the so-called hyper-
elliptic functions.
The graphs of the integrands of the integrals of the first and second
kinds are of some interest (see Fig. 4). For k = 0,
A0 ==
- k2 sin2 6 and -—: ss -
- A;2 sin2 6
both become equal to 1, and the corresponding integrals are both equal
to <p. For 0 < k < 1, the curve y = 1/A0 lies entirely above the line
•t/ = 1 and the curve y = A0 lies entirely below it. As ^> increases,
* For a detailed account see Goursat-Hedrick, Mathematical Analysis,
vol. 1, p. 226. A monograph, Elliptic Functions by H. Hancock, may also
be consulted.
§10 INFINITE SERIES 51
F(kj <p) and E(k, <p) increase continuously, F being always the larger.
As k increases, <p being fixed, the value of F(k, <p)* increases and that of
E(k, <p) decreases. Also F(k, TT) = 2K and E(k, TT) = 2E, for the curves
are symmetrical about 0 = Tr/2. If ir/2 < <p < IT, it is obvious from
the figure that
(15-7)
Moreover,
(15-8) F(k, mir + <p) - 2mK + F(k, <p),
E(k, mir + <p>) = 2mE + E(k9 <p),
where in is an integer.
Since the values of K and E, and of F(k, <p) and E(k, <p) for (p <> ir/2,
are tabulated, the relations (15-7) and (15-8) permit the evaluation of
F(k, <p) and E(k, <p) for all values of <p.
The discussion* above was restricted to values of k2 < 1. If &2 = 1,
y = A0 becomes y = |cos 0| and ?/ = 1/A0 becomes y = |sec 0|.
Consider
u = r •_==*===. =
Jo ^/(i - z2)(\. - k'W) Jo
- z2)(\. - k'W) Jo -y/i - p sin2 0
where x = sin ^>.
For a fixed value of A, (15-9) defines u = F(z) or u = Ffo>). The
function resulting from the solution of (15-9) for <p in terms of u is
called the amplitude of u and is denoted by am (u, mod &), or more
simply by <p = am u. It will be assumed that the equation u = F(<p)
can be solved for <p. Since <p — am w,
x = sin ^> = sin am u = sn u.
Moreover,
cos <p = \/l ~ ^2 s \/l — sn2 1* = en u.
Finally,
The functions sn u, en u, and dn w are called the elliptic functions.
From the definitions, it is obvious that
am (0) = 0, sn (0) = 0, en (0) = 1, dn (0) = 1;
am (— u) = —am u, sn (—u) — — sn u, en (— u) = en u, dn (— u)
= dn u.
The elliptic functions are periodic functions and in some respects
resemble the trigonometric functions. There^ exists a complete set of
* See Prob. 1, at the end of this section.
52 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §15
formulas connecting the elliptic functions analogous to the set for the
trigonometric functions.*
An interesting application of elliptic integrals to electrical
problems is found in the calculation of the magnetic flux density
in the plane of a circular loop of radius a carrying a steady
current /.
Upon applying the law of Biot and Savartf to a circular loop of
radius a, the flux density B at any point P in the plane of the
wire is given by
(15-10) B = L f "Mr,*)*
v ' 4?r Jc r*
where C is the circumference of the loop, r is the radius vector
from P to an element of arc
ds, and (r, s) is the angle
between r and this element
(Fig. 5).
If the point P is at the
center of the loop, then (r, s)
= 90°, r = a, and the integral
is easily evaluated to give
= _
4* a2 ~ 2a
a familiar result.
If, however, the point P is
not at the center, the evalua-
FlG* 5* tion of the integral is not so
easy. Consider the triangle RQS, where the side RQ — r dd
makes an angle a with ds. It is clear that ds cos a = r dQ] and,
since a = 90° — (r, s), it follows that
cos a = sin (r, s).
Hence,
, rdO
ds = ~. — -f — ^«
sin (r, s)
* See APPBL, P., and E. LACOUB, Fonctions elliptiques; PEIRCE, B. O., A
Short Table of Integrals; GBEENHILL, A. G., The Application of Elliptic
Functions.
t This formula is known to engineers as 'Ampere's formula. See, for
example, E. Bennett, Introductory Electrodynamics for Engineers. The
system of units used here is the "rational" system of units used in M. Mason
and W. Weaver, Electromagnetic Field.
§15 INFINITE SERIES 53
The substitution of this value in (15-10) yields
for the magnetic flux density at P.
Now, from triangle OQA, it is evident that
\/a2 - (r sin 0)2 = r cos 0 + A,
which, after squaring both sides and simplifying, becomes
r2 + 2rh cos 6 + (/i2 - a2) = 0.
Solving for r gives
r = -/i cos S ± \A2 cos2 0 + a2 - /i2;
and, since r is always positive, the radical must be taken with the
positive sign. Substituting this value of r in (15-11) gives
B = L r2*
4*- Jo -h
cos e + \/h2 cos2 B + a2 - h2
or, upon rationalization of the denominator,
D I C2ir-h cos 0'- Vh2 cos2 0 + a2 - h2 ,.
B = 5 Jo P"^ ^
i / r2- r2* , \
= A / o r^ ( I ^ cos 0 d6 + I A/a2 — h2 sin2 0 c?0 )-
4w(a2 - /i2) \ Jo Jo /
The first of these integrals is zero, and the second is an elliptic
integral of the second kind, so that
In
B =
A (
47r(a2
n r2v I /?2
MN \ I1 - ~2 sin2 e de
— ft2) Jo \ a2
I
T /»2
= f 2 ,2,
7r(a2 - h2) Jo
- fc2 sin2 0 dff,
where k = h/a. This integral can be evaluated for any k with
the aid of the tables of elliptic integrals.
PROBLEMS
1. Prove that
da , ^ .
/V
Jo
- I2 sin2 <p ^ o A/I - ^-2 Sin2 a
Hint: Change the variable by setting I2 sin2 ^ = sin2 a.
54 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §15
2. Plot, with the aid of Peirce's tables, F(k, ^>), where k = sin a,
using a as abscissa and F(k, <p) as ordinate. Draw 10 curves on the
same sheet of rectangular coordinate paper for <p — 0, <p = 10, <p == 20,
¥> = 30, <f> = 40, y> = 50, • • • , <p = 90.
3. Plot four curves representing F(k, p) on the same sheet of rectan-
gular coordinate paper. Use <p as abscissa and the values of k as 0,
K, \/3/2, and 1.
/•tp dp
4. Plot the integrand of I — .- - — for the values of k = 0,
Jo -v/1 - k2 sin2 <p
%, and 1. ' Use (p as abscissa. The areas under the curves give the
values of the elliptic integrals.
6. Compute the value of F(0, ir/2).
6. The major and minor axes of an elliptical arch are 200 ft. and
50 ft., respectively. Find the length of the arch. Compute the length
of the arch between the points where x — 0 and x = 25. Use Peirce's
tables.
7. Plot with the aid of Peirce's tables E(k, <p), where k = sin a.
Use a's as abscissas and E(k, <p) as ordinates. Draw 10 curves on the
same sheet of rectangular coordinate paper for (p = 0, 10, 20, • • • ,
90.
8. Plot on a sheet of rectangular coordinate paper the four curves
representing E(k, <p). Use <p as abscissa. The four curves are for k = 0,
H, \/3/2, and 1.
9. Plot the integrand of I \/l~-" k2 sin2 p dp for the values of
•/o
k — 0, H> and 1. Use v? as abscissa. Compare the result with that of
Prob. 4. What can be said about the relative magnitudes of F(k, (p)
and E(k,<p}(?
I*<P d(D
10. Show that I — . — is an elliptic integral of the first
Jo -y/i 4- k2 sin2 <p
kind.
Hint: Change the variable by setting sin <p = T tan x.
11. Show that
dx
o \/l —
cos x o \l — H sm
Hint: Set -\/cos x = cos ^>.
Note that the integral is improper but that it is easy to show its
convergence.
12. Show that
ain*gd9
__
" 2 (K -
§16 INFINITE SERIES 55
Hint: sin2 0 = j2 - ~2 (1 - k* sin2 6).
13. Show that
# ^ fl „ ^
Jo
- A;2 sin2 0
14. Find the length of one arch of the sine curve.
16. Find the length of the portion of y = sin x lying between #
and x = 2.
16. Given:
- K sin2 0
Find K and sn %K.
17. Show that I —. — » where a > 1, is an elliptic integral.
J va — cos 0
18. Show that the length of arc of an ellipse of semiaxes a and 6 is
given by
s = 4a f 2 Vl ~ e2 sin2 0 d»
Jo
= 2ira (l — Z "~ 54 g4 ""'*')> where e is the eccentricity.
16. Approximate Formulas in Applied Mathematics. It is
frequently necessary to introduce approximations in order to
make readily usable the results of mathematical investigations.
For example, an engineer seldom finds it necessary to use the
exact formula for the curvature of a curve whose equation is
y = f(x), namely,
dfy
fir2
(16-1) K = ax
since in most applications the slope dy/dx is small enough to
permit the use of the approximate formula
2
(16-2) K - |f2
Many such approximations are obtained by using the first few
terms of the Taylor's series expansion in place of the function
56 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §16
itself. Thus, the formula (16-2) is obtained from (16-1) by
neglecting all except the first term in the expansion of [1 +
(dy/dx)*\~~^ in powers of dy/dx.
I. Small Errors. The values of physical quantities determined
by experiment are subject to errors due to inaccuracies arising in
the measurements of the quantities involved. It is often neces-
sary to know the size of such errors.
Let a capillary tube contain a column of mercury. The radius
R of the tube can be determined by measuring the length L and
the weight W of the column of mercury. Let L be measured in
centimeters and W in grams. Since the density of mercury is
P = 13.6, ^
fl = J-^y = 0.153 J~
\TTpL \ L
The principal error arises in the measurement of L. Let L be the
true value, and let L' = L + € be the observed value. Then,
if R is the true value of the radius, let R' = R + 77 be the com-
puted value. The error in measuring W is negligible because of
the high accuracy of the balance. It follows that
fw IW
R = 0.153 J j- and R' = 0.153 Jj-f
or
R + TJ = 0.15
Therefore,
Since € is small compared with L, it follows that 77 is approxi-
mately given by — K R T* Clearly, c can be either positive or
Z Li
negative.
2. Crank and Connecting Rod. If one end of a straight line
PQ (see Fig. 6) is required to move on a circle, while the other
§10 , - INFINITE SERIES 57
end moves on a straight line which passes through the center of
the circle, the resulting motion is called connecting-rod motion.
This kind of motion arises in a steam engine in which one end of
the connecting rod is attached to the crank PB and therefore
moves in a circle whose radius is the length of the crank, while the
other end is attached to the crosshead and moves along a straight
line.
Let r be the length of the crank, I the length of the connecting
rod, and s the displacement of the crosshead from the position A,
D ~
FIG. 6
in which the connecting rod and crank lie in a straight line.
Then,
AB = I + r,
and
AB = AQ + QD + DB = s + I cos <p + r cos 0.
Moreover,
PI) = Z sin <p = r sin 0,
so that
sin ^> = j sin 0
and
r* r* • »*
cos a? = ^ / 1 — 77 smz 0.
\ • I2
Therefore,
Y 1 - ^ sin2 6>J
: 1 - ( 1 - ^ sin2 0J + r(l - cos 6).
0 -£-.•.)"
s + Z ( 1 - -7? sin2 0 ) + r cos 0 = I + r
or
s
If
58 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §16
be-replaced by its expansion, it follows that
fir2 1 AA2 1
s^lj^sin^ + i^j Sin40+ . . . J+r(1_cos0)
= fc sin2 6 + ^ sin4 0 +•••) + r(l - cos 0).
\&l Ol /
If r is small compared with Z, the displacement of the crosshead
is given approximately by r(l — cos 0) .
3. Surveying. In railroad surveying, it is fre-
quently useful to know the amount of difference
between the length of a circular arc and the
length of its corresponding chord.
Let r be the radius of curvature of the arc
AB (Fig. 7), and let a be the angle intercepted
by the arc. Then, if s is the length of the arc
AB and c is the length of chord AB, s = ra and c = 2r sin ~-
Since
sin x = x — ^-? + ^-j cos £ ,
where 0 < £ < x, the error in using only the first two terms of the
FIG. 7.
expansion is certainly less than
with an error less than
5!
Then,
Therefore,
1920
o-o- ,«-,„-,-_ _ —
with an error that is less than ra6/1920.
4. Vertical Motion under Earth's Attraction. Let it be required
to determine the velocity of a body of mass m that is falling from
a height s0 above the center of the earth and is subject to the
earth's attraction alone.
Let F be the attraction on the earth's surface and Ff be the
attraction at a distance h from the surface (Fig. 8). Then
§16
F =
INFINITE SERIES
kmm' , „, kmm'
59
(f
where m' is the mass of the earth, k is the gravitational constant,
and r is the radius of the earth. Hence,
F (r + h)2
F' r2
Also, let g be the acceleration at the surface of the
earth and g' be the acceleration at a distance h above
the surface, so that F = mg and F' = mg'. It
follows that
F' ~" "'
g
and, therefore,
But
a> = .
9 s2
Fia. 8.
so that
dt2
This equation can be solved for v = ds/dt by the following
device: Multiplying both members by 2 ds/dt and integrating
give
where C is the constant of integration. If the initial velocity
(ds/dt)8^8Q is zero, then C = — 2gr2/s0 and hence
But s = r + A and ds/eS = v, so that the equation becomes
( — J-r
\r +
This formula can be used to calculate the terminal velocity
(i.e., the velocity at the earth's surface) when the body is released
60 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §16
from any height. Thus, setting h = 0 gives
(16-3) ^ = 2^(1-1).
Upon denoting by ho the initial height above the earth's surface,
so that SQ = r + ho, (16-3) can be written as
or
(16-4) v2 = 2gr
Now , , = ( 1 H — - J ; and if — < 1, then series expansion
is permissible, so that
I .f If/Q
^ r + ho = 7 "
Hence, if ho/r < 1, (16-4) can be replaced by
Moreover, if &0 is very small compared with r, then the powers
of ho/r higher than the first Can be neglected* and
v2 = 2ghQ,
which is the familiar formula for the terminal velocity of a body
falling freely from a height ho that is not too great.
It follows from (16-3) that the square of the terminal velocity
will be less than 20r2(l/r) = 2gr. Moreover, for large values of
s<j the terminal velocity will be very close to \/2gr. Accordingly,
if a body falls from a very great distance it would attain a ter-
minal velocity (air resistance being neglected) of approximately
\/2gr = 6.95 miles per second.
The results stated in the last paragraph may receive a different
interpretation. Suppose a body were projected outward from
the earth's surface with a velocity of more than \/2gr = 6.95
miles per second. The previous discussion shows that, if air
* Since the series is alternating, the error will be less than 2gr(hQ/r)*.
§16 INFINITE SERIES 61
resistance is neglected, the body would travel an infinite dis-
tance. This velocity is called the critical velocity or the velocity
of escape.
It may be recalled that the earth's rotation exerts a centrifugal
force on a particle which is falling toward the earth and that this
force diminishes the effect of the force due to the earth's attrac-
tion. For a particle of mass m on the surface of the earth at the
equator, this centrifugal force is
mv2 mw2r2 0 mq ,
_ = ___ = ww-r = ___ dynes,
where o> = 0.00007292 radian per second is the angular velocity
of the earth, r = 6,370,284 m., and g = 980 cm. per second per
second. At a distance s from the center of the earth, this force is
o mas
m»s = m-'
But the earth's attraction at this distance is F = mgf. Since
9f = 0r2M
_ mgr*
* ~ s2 '
If the particle is to be in equilibrium,
mgs _ mgr2
289r ~ ~s2~'
so that
s3 = 289r3 or s = 6.6r = 26,000 miles approx.
Thus, if all other forces are neglected, a particle would be in
equilibrium at approximately 22,000 miles above the earth's
surface. This gives a very rough approximation to the extent
of the earth's atmosphere. The actual thickness of the atmos-
pheric layer is supposed to be considerably smaller.
PROBLEMS
1. The mass of the moon is nearly one-eighty-first that of the earth,
and its radius is approximately three-elevenths that of the earth.
Determine the velocity of escape for a body projected from the moon.
Acceleration of gravity on the surface of the moon is one-sixth that on
the surface of the earth.
62 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §16
2. Show that the time required for a body to reach the surface of the
earth in Illustration 4, Sec. 16, is
Hint:
ds
3. If the earth is considered as a homogeneous sphere at rest, then
the force of attraction on a particle within the sphere can be shown to
be proportional to the distance of the particle from the center. Let a
hole be bored through the center of the earth, the air exhausted, and
a stone released from rest at the surface of the earth. Show that the
velocity of the stone at the center of the earth is about 5 miles per
second. •
Hint:
d*s mg
"&---•>
where 8 is the distance of the stone from the center of the earth and r
is the radius of the earth.
CHAPTER III
SOLUTION OF EQUATIONS
Students of engineering, physics, chemistry, and other sciences
meet the problem of the solution of equations at every stage of
their work. This chapter gives a brief outline of some of the
algebraic, graphical, and numerical methods of obtaining the
real roots of equations with real coefficients, of types that occur
frequently in the applied sciences. It also contains a short
summary of those parts of the theory of determinants and the
theory of matrices that are immediately applicable to the solution
of systems of linear equations.
26. Graphical Solutions. The subject of the solution of equa-
tions will be introduced by considering a simple problem that
any engineer may be called upon to solve.
It is required to design a hollow cast-iron sphere, 1 in. in
thickness, that will just float in water. It is assumed that the
air in the cavity is completely exhausted. The specific gravity
of cast iron will be denoted by p, for convenience.
By the law of Archimedes, the weight of the sphere must
equal the weight of the displaced water. This gives the con-
dition on the radius of the sphere, namely,
- (x - I)3].
Simplifying gives
(25-1) x3 - Spx* + 3pz - p = 0.
It will be convenient to remove the second-degree term in (25-1).
To accomplish this, let x = y + k, giving
Zyk* + & - 3P(i/2 + 2yk + fc») + 3P(y + k) - /> - 0,
or
y8 -f (3fc - 3p)i/2 + (3fc2 - Qpk + 3p)y + W - 3pfc2 4- 3pfc - p - 0.
Choosing k = p makes the equation reduce to
(25-2) 2/8 + (3p ~ 3P2)2/ - 2P3 + 3p2 - p = 0.
83
84 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §26
For cast iron, p = 7.5, and (25-2) becomes
(25-3) y* - 146.257/ - 682.5 = 0.
If (25-3) is solved, the solution of (25-1) is also determined, since
x = y + 7.5.
A graphical method of solution will be used. The solution
of (25-3) is equivalent to the simultaneous solution of the system
(25-4) (* = 2/3>
v ' \z = 146.25*/ + 682.5.
The accompanying figure (Fig. 16) represents the graphs of the
two functions of (25-4); since they inter-
sect at y = 14.0, this value gives an
approximate solution of (25-3). The cor-
responding solution of (25-1) is x = 21.5.
From the graph, it is clear that there is
only one real solution of (25-4) and hence
6825,
of (25-3).
M0 This graphical method can be applied to
Fia- 16> any cubic equation. The general fourth-
degree equation (quartic) can also be reduced to a form that is
convenient for graphical methods of solution.
Consider the quartic
x4 + ax3 + bx* + ex + d = 0.
Let x = y + k, as in the cubic equation. This substitution
gives
+ ?/(4fc3 + 3ak2 + 2bk + c) + fc4 + ak* + bk2 + ck + d = 0.
In order to remove the term in y3, choose k = — -T- This reduces
the equation to the form
2/4 + Ay* + By + C = 0.
If A > 0, the further transformation y — \/A z is made, and the
equation is reduced to
or
u4 + AW + B VA z + C = 0,
s4 + ** + P* + q = 0.
§26 SOLUTION OF EQUATIONS 85
The solutions of this equation are the same as the solutions of
the simultaneous system
u = z4 + z2,
u = —pz — q.
The graphs of these two functions are easily plotted, and the
solutions can be read from the graph. In case A < 0, the
transformation would be y = \/ — A z, which leads to the equation
• z4 - z2 + pz + q = 0
and the graphical solution of the system
u =
w = —pz — q.
This method of solution for the real roots of an equation is
also applicable to many transcendental equations. In order to
solve
Ax — B sin x = 0,
write it as
ax — sin x — 0,
and plot the curves of the simultaneous system
i y = sin x,
y = ox.
Similarly, the equation
a* ~ X2 = o
can be solved graphically by plotting the curves of the equivalent
simultaneous system
y = o*,
y = x2.
PROBLEMS
1. Solve graphically
(a) 2* - x* = 0,
(5) a?4 - x - 1 = 0,
(c) x6 - Z - 0.5 = 0,
(d) e* + x = 0.
2. Find, graphically, the root of
tan x — x = 0
nearest %TT.
86 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §26
26. Algebraic Solution of Cubic. The graphical method of solution
is perfectly general, but its accuracy depends upon the accurate con-
struction of the graphs of the equations in the simultaneous' systems.
This is often extremely laborious and, at most, yields only an approxi-
mate value of the roots.
In the case of the linear equation ax + b — 0, where a 5^- 0, the solu-
tion is x = —b/a. For the quadratic equation ax2 + bx + c = 0,
— b ± \/62 — 4ac
where a 7* 0, there are two solutions given byx
The question naturally arises as to the possibility of obtaining expres-
sions for the roots of ajgebraic equations of degree higher than 2.
This section will be devoted to a derivation of the -solutions of the
general cubic equation «.
dox* + aix2 + aw + a3 = 0, * a0 5^ 0.
Dividing through by ao gives
(26-1) z3 + bx2+^cx + d = 0,
and the x2 term can be removed by making the change of variable
6
f — tii .. — .
* ~ J 3
The resulting equation is
(26-2) y» + py + q = 0,
where
b2
p = c-T
and
. be , 263
«-d--3 + 27-
In order to solve (26-2), assume that
(26-3) y = A + B,
so that
?/ = A3 + B* + 3AB(A + B).
Substitute in this last equation for A + B, from (26-3), and there is
obtained the equation
(26-4) ?/ - 3ABy - (A3 + £3) = 0.
A comparison of (26-4) with (26-2) shows that
SAB = -p and A3 + B3 = -3,
SOLUTION OF EQUATIONS
87
or
(26-5)
A8£3 = - •%= and A3 + B9 = -$.
If B3 is eliminated by substituting from the second of Eqs. (26-5) into
the first, there appears the quadratic equation in A8,
whose roots are
-q±
The solution for Bz yields precisely the same values. However, in
order to satisfy Eq. (26-5), choose*
(26-6)
B* =
If the values of y are to be determined from (26-3), it is necessary to
find the cube roots of A3 and j53. Recall that if xz = a3, then the solu-
tions for x are given by a, coa, and o>2a, where co = — ^ H — o~ i
o2 = — o -- o~ ^ are
complex roots of unity. Hence, if one cube
root of A3 be denoted by a and one cube root of B3 by £, the cube roots
of Az are
a, coa, and w2a,
whereas those of J53 are
|9, cop, and co20.
It would appear that there are nine choices for «/, but it should be
remembered that the values must be paired so that SAB = — p. The
only pairs that satisfy this condition are a and /3, coa and co'2j0, and o>2a
and wjS, Hence, the values of y are
(26-7) yi = a + ft 2/2 = coa + a>2ft 2/3 = co2a + coft
where
and
* The opposite choice for the values of A8 and J53 simply interchanges
their role in what follows.
88 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §26
The solutions of (26-1) can be obtained from the values given in
(26-7) by recalling that x = y - 6/3.
The expressions for a and 0 are quite complicated, and when the
quantity under the square-root sign has a negative value the values of
a and ft cannot, in general, be determined. This is the so-called
irreducible case of the cubic, which can, however, be solved by using a
trigonometric method. This method will be
described later in the section, but first it is im-
portant to find a criterion that will determine
in advance which method should be used.
In order to determine the character of the
.y roots of (26-2), whose coefficients are assumed
to be real, consider the function
O
f(y) == y3 + py + q
FIG. 17. and its maximum and minimum values.
f(y) = 3i/2 + p,
Since
it appears that, if p > 0, then f'(y) is always positive and f(y) is an
increasing function. In this case the graph of f(y) has the form shown
in Fig. 17, and there is evidently only one real value for which f(y) = 0.
If p < 0, however, f(y) is zero when y = ± \/-p/3. Since
f"(y) = 6i/, it follows that y = + V- P/3 gives a minimum value to
f(y), whereas y = — V' — p/S furnishes a maximum value. The cor-
responding values of f(y) are
ffyr
and
The graph of f(y) will have the
appearance of one of the curves in
Fig. 18.
It is evident that f(y) = 0 will
have only one real root if the graph
of f(y) has the appearance shown by (1) or (5), that is, if the maximum
and minimum values of f(y) are of the same sign. Hence,
or
FIG. 18.
0,
§26 SOLUTION OF EQUATIONS .89
is the condition that (26-2) have only one real root. It may be observed
that this condition is automatically satisfied if p ^ 0. It should be
noted that, if p = 0, Eq. (26-2) reduces to y* + q = 0, which obviously
has only one real root.
If (26-2) has three real and distinct roots, then the graph of f(y) must
have the appearance shown in (3), and it follows that the maximum
and minimum values must be of opposite sign. Hence,
q2 + «7 P3 < 0
is the condition for three real and unequal roots.
If <f + $iiP* = 0, either the maximum or the minimum value of
f(y) must be zero [see (2) and (4)], and (26-2) will have three real roots,
of which two will be equal (a so-called double root).
The expression
(26-8) A s -27g2 - 4p3
is called the discriminant of the cubic equation (26-2), for its value
determines the character of the roots of the equation. The discriminant
for (26-1), obtained by replacing p and q in (26-8) by their values in
terms of 6, c, and d, is
(26-9) A s ISbcd - 463d + &2c2 '- 4c3 - 27d\
It may be worth noting that the discriminant of any algebraic equa-
tion, with leading coefficient unity, is the product of the squares of
the differences of the roots taken two at a time. Inasmuch as
(xi - X2)*(x2 - X3y(x3 - XiY = (yi - 2/2)2(?/2 - 2/3)2(2/3 - 2/i)2,
the discriminant has the same value for (26-1) and (26-2).
In view of the definition of A, it follows that
if A < 0, one root is real and two are complex;
if A = 0, all the roots are real and two are equal;
if A > 0, the three roots are real and unequal.
Example. Consider the cubic equation
x* + 3x* + 9z - 1 = 0.
From (26-9), it follows that A = —2592, and hence there will be one real
root and two complex roots. Setting x = y — 1 yields the reduced cubic
2/* + 6y - 8 = 0,
and substituting p = 6 and q = —8 in (26-6) gives A3 =4 +2 ^/Q and
B3 = 4 — 2 V6. Therefore, the solutions for y are
^4 + 2 \/6 4- ^4 - 2 \/6, « ^4 + 2 \/6 + w2 ^4 - 2 v/6,
and w2 ^4 +2\/6 + w ^4-2 \/6.
90. MATHEMATICS FOR ENGINEERS AND PHYSICISTS §26
The solutions of the original equation can now be obtained by recalling that
gW y - 1.
The discussion of the solution of the cubic* equation "will be con-
cluded by giving the derivation of the expressions for the roots in the
case where the roots are real and unequal (that is, when A = — 27 q2
- 4p3 > 0).
Let
— ~ = r cos 0
Jj
and
- + -"•*
Then*
(A Q\
cos - + i sin - )
o o/
and
(n nv
cos - — i sin - V
If it is noted that
and
o
2
2?r . . . 2ir
= cos ~ + i sm ~
o o
27T . . 27T
o
co2 = cos -r- — i sm
3 3
it is easily checked that the expressions for
2/i == a + 0, 7/2 = wo: + O)2j8, 2/3
become
(26-10) ?/i - 2rH cos 1 ya = 2rH cos ^
o
COS
Since
'-•v-fr
and
cos 8 = —
the values of 2/1, 2/2, and 2/3 can be obtained directly from the coeffi-
cients of (26-2) or (26-1).
* By De Moivre's theorem (cos 0 -f- i sin 0)" = cos nO + i sin n0.
§26 SOLUTION OF EQUATIONS 91
Example. Determine the real roots of
x* - 3ic2 + 3 » 0.
Here
A fc= -4(-27)(3) -27(9) > 0,
and the roots are all real and unequal. Since p — — 3 and q *= 1, it follows
that r - 1 and cos 0 = — >£. Hence,
and
*27r Sir n STT
i/i=2 cos — > 1/2=2 cos — > 2/3 = 2 cos -r-
• y y «7
The solutions of the general quartic equation
z4 + kr3 + ex* + cfc + e = 0
can be found, but the methods of obtaining the expressions for the
roots depend upon the solution of an auxiliary cubic equation. More-
over, these expressions are, in general, so involved that they are prac-
tically useless for computation.* It has been shown that the ordinary
operations of algebra are, in general, insufficient for the purpose of
obtaining ex'act solutions of algebraic equations of degree higher than 4.
However, it is possible to obtain the expressions for the solutions of the
general equation of the fifth degree with the aid of elliptic integrals.
The reader should not confuse the problem of obtaining expressions
for the exact solutions of the general algebraic equation with that of
calculating numerical approximations to the roots of specific equations
which have numerical coefficients. The latter problem will be dis-
cussed in Sees. 28 and 29, and it will be shown that the real roots of
such equations can be computed to any desired degree of accuracy.
Moreover, if the roots are rational they can always be determined
exactly.
PROBLEMS
Determine the roots of the following equations:
(a) ?/3 - 2y - 1 = 0;
(6) 7/ - 146.25y - 682.5 = 0;
(c) xs - x2 - 5x - 3 = 0;
(d) x* - 2x* - x + 2 = 0;
(e) x* - 6z2 + 6z - 2 = 0;
(/) x3 + 6z2 + 3x + 18 = 0;
(0) 2x* + 3z2 + 3s + 1 = 0.
* See DICKSON, L. E., First Course in Theory of Equations, pp. 50-54;
BURNSIDE, W. S., and A. W. PANTON, Theory of Equations, vol. 1, pp. 121-142.
92 MATHEMATICS FOR ENGINEERS AND PHYSICISTS &27
27. Some Algebraic Theorems. The student of any applied
science is usually interested in obtaining numerical values,
correct to a certain number of decimal places, for the roots of
equations. Unless the roots are rational, the expressions for the
exact roots, provided that they can be found at all, are usually
complicated and the process of determining numerical values
from them is tedious. Accordingly, it is distinctly useful to
consider other methods of finding these numerical values.
Plorner's method, Newton's method, and the method of inter-
polation are the ones most frequently used; they will be dis-
cussed in Sees. 28 and 29. However, all thcksc methods arc
based on the assumption that a root has first been isolated, that
is, that there have been determined two values of the variable
such that between them lies one and only one root. In many
practical problems the physical setup is a guide in this isolation
process. This section contains a review of some theorems* from
the theory of equations that provide preliminary information as
to the character and location of the roots.
THEOREM 1 (Fundamental Theorem of Algebra.) Every
algebraic equation
f(x) == a«xn + air"-1 + • • + an-iz + an = 0
has a root.
It should be noted that this theorem does not hold for non-
algebraic equations. For example, the equation ex — 0 has
no root.
THEOREM 2. (Remainder Theorem.) // the polynomial
f(x) E= aQxn + aixn~l +••••+ an-ix + an
is divided by x — I) until the remainder is independent of x, then
this remainder has the value f(b).
THEOREM 3. (Factor Theorem.) // f(b) = 0, then x — b is
a factor of the polynomial /(x) and b is a root off(x) = 0.
This theorem follows directly from Theorem 2. In many
cases the easiest way to compute the value of f(b) is to perform
the division of f(x) by x — b. This is a particularly useful
* Those students who are not already familiar with these theorems and
their proofs will benefit by referring to H. B. Fine, College Algebra, pp.
425-453, and L. E. Dickson, First Course in the Theory of Equations, Chap.
II.
§27 SOLUTION OF EQUATIONS 93
method when the factor theorem is being used for the purpose of
determining the roots of f(x) = 0. For if x — b is a factor of
/Or), it follows that f(x) = (# — &) g(x), where g(x) is a poly-
nomial of degree one less than that of /(#). Obviously the roots
of g(x) = 0 will be the remaining roots of f(jr) — 0, so that only
g(x) = 0 need be considered in attempting to find these roots.
Moreover, when /(>) is divided by x — b the quotient is g(x).
If synthetic division is used, the computation is usually quite
simple.
Example. If }(x) = z3 + 2x2 + 2x + 1 is divided by x + 1, the
quotient is x2 + x + 1 and the remainder is zero. Hence, x — —1 is
a root of f(x) — 0 and the remaining roots are determined by solving
X2 + x + i = o.
THEOREM 4. Every algebraic equation of degree n has exactly
n roots if a root of multiplicity m is counted as m roots.
A root b of f(x) = 0 is said to be a root of multiplicity m if
(x — b)m is a factor of /(x) but (x —• b)m+1 is not a factor of f(x).
It follows from Theorems 2 and 4 that the polynomial of
degree n can be factored into n linear factors, so that
f(x) = a<>xn + aixn~l + • ' • + dn-ix + an
= a0(x — xi)(x — x2) • • • (x — xn).
THEOREM 5. '//
f(x) = a0xn + aixn~l + • • • + an^x + an
has integral coefficients and if f(x) = 0 has the rational root b/c,
where b and c are integers without a common divisor, then b is an
exact divisor of an and c is an exact divisor of a0.
Example. Consider the equation
f(x) = 2x* + x2 + x - 1 = 0.
The only possible rational roots are ± 1 and ± J£. Since /(I) = 3,
/(-I) = -3, f(-lA] = -%, and/(K) = 0, it follows that K is the
only rational root. As a matter of fact, if f(x) is divided by x — % the
quotient is 2x2 + 2x + 2 whose factors are 2, x — co, and x — o>2,
where w and co2 are the complex roots of unity.*
THEOREM 6. Given f(x) = a;n + ^i^""1 + • • • + an~ix + an
= 0. Iff(a) andf(b) are of opposite sign, then there exists at least
* See Sec. 26 and the example following Theorem 9 of this section.
94 MATHEMATICS FOR ENGINEERS AND PHYSICISTS ^27
one root off(x) = 0 between a and b. Moreover, the number of such
roots is odd.
Graphically this means that y = f(x) must cross the :c-axis
an odd number of times between a and b.
Example. If f(x) s 8x* - I2x2 - 2x + 3 = 0,
/(-I) = -15, /(O) = 3; /(I) - -3, /(2) = 15.
Since /( — I) is negative and /(O) is positive, there is at least one root
between —1 and 0. Similarly, there is a root between 0 and 1, and
another between 1 and 2.
THEOREM 7. (Descartes' Rule of Signs.) The number of
positive real roots of an algebraic equation f(x) — 0 with real coeffi-
cients is either equal to the number of variations in sign of f(x) or less
than that number by a positive even integer. The number of negative
real roots of f(x) = 0 is either equal to the number of variations in
sign of f( — x) or less than that number by a positive even integer.
Example. f(x) = 8#3 — I2x2 — 2x + 3 has two changes in sign, and
therefore there are either two or no positive roots of f(x) = 0. Also,
f( — x) s — &c3 — I2x2 + 2x + 3 has only one change in sign, and/(#)
must have one negative root.
THEOREM 8. Every algebraic equation of odd degree, with real
coefficients, and leading coefficient positive has at least one real root
whose sign is opposite to that of the constant term.
Example. Since f(x) s= 8x* — 12z2 — 2x + 3 = 0 is of odd degree
and the constant term is positive, it follows that there must be at least
one negative root.
THEOREM 9. // an algebraic equation f(x) = 0 with real coeffi-
cients has a root a + bi, where b 9^ 0, and a and b are real, it also has
the root a — bi.
Example. Thus, x* — 1 = 0 has the root — M + K \/3 1, and there-
fore it has the root — % — J£ \/3 i. This theorem states that imaginary
roots always occur in pairs.
PROBLEMS
1. Find all the roots of the following equations:
(a) xs + 2x* - 4z - 8 = 0;
(6) 2x* -x2 - 5x - 2 = 0;
(c) 4z4 + 4z3 + 3z2 - x - 1 = 0;
(d) 2z4 - 3z3 - 3x - 2 « 0.
§28 SOLUTION OF EQUATIONS 95
2. Isolate the roots of the following equations between consecutive
integers :
(a) x3 - 2x2 - x + 1 = 0;
(6) 2s3 + 4z2 - 2x - 3 = 0;
(c) x3 -f 5z2 + to + 1 = 0;
(d) x4 - 5z2 + 3 = 0.
28. Hornet's Method. Many readers are already familiar with
Horncr's method of determining the value, to any desired number of
decimal places, of the real roots of algebraic equations. However, the
development given here is somewhat different from that used in the
texts on algebra, in that it depends on Taylor's scries expansion.
Suppose that the equation is
(28-1) f(x) =s a0xn + ttix"-1 + • • • + (in^x + an = 0
and that it is known that the equation has a root between c and c + 1 ,
where c is an integer. If f(x) is expanded in Taylor's series in powers of
x — c, there will result* a polynomial in x — r, namely,
i"(c\
/« +f'(e)(x - c) + J- (x - c)« + • • •
Now, let x — c = Xi and — p~ == Aft_r. Then (28-1) is replaced by
(28-2) /t(xi) - An + An-iX! + - • • + AiX!-1 + Aox,« - 0.
Since (28-1) had a root between c and c + 1 and since Xi = x — c,
it is evident that (28-2) has a root between 0 and 1. By the use of
Theorem 6, Sec 27, this root can be isolated between d and d + 0.1,
where d has the form a/ 10 and 0 < a < 9. Moreover, /i(xj = f(x\ -f c) ;
and it follows, that, if f\ has a root between d and d -f- 0.1, then / has a
root between c + d and c + d + 0.1. It should be noted that c may
be negative but that d will always be positive or zero.
The function /i(^i) can be expanded in Taylor's series in powers 'of
x\ — d] and, if #2 = #1 — d, there will be obtained an equation
/2(a?2) = Bn + Bn^xt + • • ' + JW-1 + Boxf = 0.
But /i (0*1) = 0 had a loot between d and d + 0.1 ; and since xz = x\ — d,
U follows that/2(x2) = 0 will have a root between 0 and 0.1.
This process can be continued as long as desired, each step deter-
mining another decimal place of the root of the original equation (28-1).
* Since /(#) is a polynomial of the nth degree, the derivatives of ordor
higher than n are all zero.
96 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §28
The solution of a specific equation may help to clarify the procedure.
Let it be required to find the values of the real roots of the equation
F(x) ss x4 + x* - 3z2 - 6x - 3 = 0.
Since there is only one variation in sign, F(x) has at most one positive
root. F(—x) has three variations, and so there will be at most three
negative roots. The only possibilities for rational roots are ± 1 and
±3. Since F( — 1) = 0, it follows that x = —1 is a root. Moreover,
if F(x) is divided by x + 1, the quotient is f(x) = x3 — 3x — 3. Hence,
the remaining roots of F(x) = 0 are the three roots of
/(x) = xs - 3x - 3 = 0.
It is easily checked that /(x) = 0 has no rational roots. Also,
A = 108 — 243, so that'there is only one real root which, since jf(2) = — 1
and /(3) = 15, must lie between 2 and 3. Therefore, f(x) will be
expanded in powers of x — 2. Since
/(x) = x*- 3* -3, /(2) = -1,
f(x) = 3x2 - 3, /' (2) - 9,
/"(x) = Gx, /"(2) = 12,
/'"(x) = 6, /'"(2) = 6,
the expansion becomes
/(x) = -1 + 9(x - 2) + ly (a; - 2)2 + |j (x -
Replacing a? — 2 by x\ gives
/1(a?l) s -1 + 9si + 6xi2 + xi3 = 0.
Since the real root of this equation lies between 0 and 1, the #i2 and Xi3
terms do not contribute very much to the value of f\(xi). Hence, a
first approximation to the root can be obtained by setting 9#i — 1 = 0.
This gives x\ — % =0.111 • • • , and suggests that the root probably
lies between 0.1 and 0.2. It is easy to show that /i (0.1) = —0.039 and
/i (0.2) = 1.048; there is thus a root between 0.1 and 0.2, and it is
evidently closer to 0.1. Therefore, f(x) = 0 has a root between 2.1
and 2.2.
Expanding /i Co; i) in powers of Xi — 0.1 gives
jfi(si) = -0.039 + 10.23(*i - 0.1) + !|y (xi - O.I)2 + » (x, - 0.1)',
and replacing x\ — 0.1 by x2 yields
/2(*2) = -0.039 + 10.23^2 + 6.3X22 + $2* - 0.
Now 10.23x2 — 0.039 = 0 gives the approximation x± = 0.0038, and
testing 0.003 and 0.004 reveals that /2(0.003) = -0.008253273 and
§29 SOLUTION OF EQUATIONS 97
/2(0.004) = + 0.002020864. Thus, the root lies between 0.003 and 0.004
and is closer to 0.004. If it is desired to determine the root of f(x) = 0
to three decimal places only, this value will be 2.104. If more decimal
places are desired, the process can be continued. It should be noted
that in each succeeding step the terms of the second and third degree
contribute less, so that the linear approximation becomes better.
PROBLEMS
1. Apply Horner's method to find the cube root of 25, correct to three
decimal places
2. Determine the real roots of £3 — 2x — 1 = 0 by Horner's method.
3. Determine the root of x4 + x3 — 7x2 — 3 + 5 = 0, which lies
bet ween » 2 and 3.
4. Determine the real root of 2x3 — '3x2 + # — 1 — 0.
5. Determine the roots of x3 — 3#2 + 3 = 0.
6. Find, correct to three decimal places, the value of the root oi
x5 + 3#3 — 2x2 + x + 1 =0, which lies between — 1 and 0.
7. A sphere 2 ft. in diameter is formed of wood whose specific gravity
is M. Find to three significant figures the depth h to which the sphere
will sink in water. The volume of a spherical segment isTT/i2 ( r — ~ Y
The volume of the submerged segment is equal to the volume of the
displaced water, which must weigh as much as the sphere. Since water
weighs 62 5 Ib. per cubic foot,
and, since r — 1,
/i3 - 3A2 + % = 0.
29. Newton's Method. Horncr's method of obtaining a
numerical solution of an equation is probably the most useful
scheme for solving algebraic equations, but
it is not applicable to trigonometric, ex-
ponential, or logarithmic equations. A
method applicable to these types as well as
to algebraic equations was developed by Sir
Isaac Newton sometime before 1676.
Newton applied his method to an alge-
braic equation, but it will be introduced
here in the solution of a problem involving FlG 19-
a trigonometric function.,
Let it be required to find the angle subtended at the center of
a circle by an arc whose length is double the length of its chord
98 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §29
(Fig. 19). Let the arc BCA be an arc of length 2BA. Let 2x
be the angle (measured in radians) subtended at the center of the
circle. Then, arc BCA = 2xr and BA = 2 DA = 2r sin x. If
arc BCA = 2BA, then 2xr = 4r sin x, or rr — 2 sin # = 0.
The graphical solution of equations
of this type was discussed in Sec. 25.
A first approximation can be obtained
by graphical means. If y = x and
?/ = 2 sin x are plotted, it appears
from the graph (Fig. 20) that they
intersect for x lying between 108° and
109°, or, expressing this in radians,
x -?$//? x
FIG. 20.
1.8850 < x < 1.9024.
If xi = 1.8850 be chosen as the first approximation, the question
of improving this value will be discussed first from the following
graphical considerations.
If the part of the curve y = x — 2 sin x in the vicinity of the
root be drawn on a large scale, it will have the appearance shown
in Fig. 21. It is clear from the graph that adding to x\ the
FIG. 21.
distance AE, cut off by the tangent line to the curve at x\ =
'1.8850, will give a value x% which is a better approximation to the
actual root XQ. But AE is the subtangent at Xi and is equal to
where /(x) = x — 2 sin x. Thus,*
Xz « Xl _ /fr)..
* See, in this connection, Prob. 8, at the end of this section.
§29 SOLUTION OF EQUATIONS 99
Similarly, upon using £2 as the second approximation and
observing that — ff \ . is the subtangent EF, the third approxi-
/(**)
mation is found to bo
and in general the nth approximation xn is given by
(29-1) xn = Xn-.! - (r^r~y (n = 2, 3, • • • )•
Since x\ = 1.885, the formula gives
_ _ f(x*) — . _ -ri ~~ ^ s*n Xl
f'(%i) l 1—2 cos 0:1
= 1.8850 - ^-^2 = 1.8956.
In a similar way,
L895(i " 2 sin h895G - 1 8955
1.8955.
f(x)
It follows that the angle subtended by the arc is 3.7910 radians.
The use of Newton's method requires some preliminary
examination of the equation. It may happen that the equation
is of such a character that the second approximation to XQ will be
worse than the first. A careful examination of the following
sketches of four types of functions,
sketched in the vicinity of their
roots, reveals the fact that some
care must be exercised in applying
Newton's method. For all four
figures, it is assumed that x0 has
been isolated between Xi and x[.
The graphical interpretation of the
f(xi)
correction —
as the subtan-
Fio. 22.
gent must be kept in mind throughout this discussion. If x\ is
used as the first approximation, then x2 will be obtained as the
second approximation by using Newton's method; if x( is used,
then £2 will be obtained.
In Fig. 22, both x% and x'% are closer to x0 than x\ or x{. In this
case the method would work regardless of which value is chosen
100 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §2$
as the first approximation. In Fig. 23, x2 is better than the
first approximation xi, but x'2 is worse than x{. It appears from
FIG. 23.
the figure that this occurs because the curve is concave down
between xi and x[, and hence f"(x) < 0, whereas f(x i) < 0 and
f(x{) > 0. A similar situa-
tion would obtain if the curve
is concave up, so that/"(#)
> 0 (Fig. 24). The reader
will readily convince himself
from an inspection of Fig 23
>x that caution must be ex-
ercised in the choice of the
first approximation if the
curve has a maximum (or a
minimum) in the vicinity of XQ.
If the curve has the appearance indicated in Fig. 25, then it is
evident that the choice of either x\ or x\ as the first approximation
will yield a second approxima-
tion which is worse than the
fir§t one. This is due to the
fact that the curve has a point
FIG. 24.
of inflection between x\ and
r'
Xi.
From the foregoing discus-
sion, it is apparent that New-
ton's method should not be
applied before making an investigation of the behavior of the first
and second derivatives of f(x) in the vicinity of the root. The
FIG. 25.
I
'•^
FIG. 26.
§29 SOLUTION OF EQUATIONS 101
conclusions drawn from this discussion can be summarized in the
following practical rule for determining the choice of the first
approximation: Iff'(x) andf"(x) do not vanish in the given interval
(xi, x{) and if the signs off(xi) and f(x{} are opposite, then the first
approximation should be chosen as that one of the two end points for
which f(x) andf"(x) have the same sign.
It can be proved* that if the single-valued continuous function
f(x) is of such a nature that /(
= 0 has only one real root in foo
(xi, x{) and both/'(s) and /"(a;)
are continuous and do not
vanish in (xi, x'^)y then repeated
applications of Newton's
method will determine the value
of the root of f(x) = 0 to any
desired number of decimal
places.
The cases to which Newton's
method does not apply can be
treated by a method of interpolation (regula falsi) that is appli-
cable to any equation.
Let x be the value of x for which the chord AB intersects the
x-axis. From similar triangles (Fig. 26),
x — Xi x( — x
Solving for x gives
. _
The value x is clearly a better approximation than either xi
or x(.
PROBLEMS
1. Solve Prob. 7, Sec. 28, by Newton's method. Also, apply the
method of interpolation.
2. Determine the angle subtended at the center of a circle by a chord
which cuts off a segment whose area is one-quarter of that of the circle.
3. Find the roots of e* — 4x — 0, correct to four decimal places.
4. Solve x — cos x — 0.
* See WEBER, H., Algebra, 2d ed. vol. 1, pp. 380-382; COATE, G. T., On the
Convergence of Newton's Method of Approximation, Amer. Math. Monthly,
vol. 44, pp. 464-466, 1937.
102 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §30
6. Solve x = tan x in the vicinity of x = %TT.
6. Solve x + e* = 0.
7. Solve x* - x - 1 = 0.
8. Show with the aid of Taylor's series that, if x = x t is an approxi-
mate root of f(x) — 0, then the nth approximation is, in general,
determined from the formula (29-1).
f(n)(r,\
Hint:f(x) = /(*0 + /'(*i)(* -*!)+•••+ J—~rL (x - *i)» + • • ;
and if f(xz) == 0, then
30. Determinants of the Second and Third Order. The solu-
tion of systems of linear equations involves the determination of
the particular values of two or more variables that will satisfy
simultaneously a set of equations in those variables. Since the
discussion is simplified by using certain properties of deter-'
minants and matrices, the remainder of this chapter is devoted
to some elementary theory of determinants and matrices and its
application to the solution of systems of linear equations.
Consider first a system composed of two linear equations in
two unknowns, namely,
(30-1) • ( aiX + ?I2/ = J"
\a2x + b2y = k>2.
If y is eliminated between these two equations, there is obtained
the equation
(30-2) (aibz — a2bi)x = kib2 — fobij
and if x is eliminated, there results
(30-3) (aj62 — a2bi)y = ai&2 — Q^i.
If the expression aj)2 — a2bi is not zero, the two equations
(30-2) and (30-3) can be solved to give values for x and y. That
the values so obtained are actually the solutions of the system
(30-1) can be verified by substitution in Kqs. (30-1).
The expression ai62 — azbi occurs as the coefficient for both
x and y. Denote it by the symbol
(30-4) 2
#2 02
This symbol is called a determinant of the second order. It is
also called the determinant of the coefficients of the system
§30
SOLUTION OF EQUATIONS
103
(30-1), for the elements of its first column are the coefficients of
x and the elements of its second column are the coefficients of y.
Using this notation, (30-2) and (30-3) become
(30-5)
a2 6;
bi
&2 o2
V =
The definition (30-4) provides the method of evaluating the
symbol. If
D -
the unique solution of (30-1) can be written as
:fci
x —
D
y = —
If Z) = 0, ai62 = fl2&i or ai/a2 = 61/62. But if the correspond-
ing coefficients of the two equations are proportional, the two lines,
whose equations are givon by (30-1), arc parallel vif a\/ai ^
fci//c2) or coincident (if 0,1/0,% = 61/62 = fci//c2). In the first
case, the determinants appearing as the right-hand members of
the equations in (30-5) will be different from zero and there will
be no solution for x and y. In the second case, these deter-
minants, as well as D, are zero and any pair of values £, y that
satisfies one equation of the system will satisfy the other equa-
tion, also.
Example 1. For the system
2x - By = -4
3x - y = 1,
-4 -3
1 -1
D =
2 -3
= -2
2 -4
3 1
7
=* 2.
Example 2. For the system
but
2 _
6 "
2 -3
6 -9
= 0,
The two lines whose equations are given are parallel.
104 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §30
Example 3. For the system
,, 2-3
2x - 3y = 4
QX - 9t, = 12, j
2 -3 4
6 ~ -9 ~~ 12
-9
= 0,
The two lines are coincident.
Consider next the system of three linear equations in three
unknowns,
+ b\y + CiZ =
(30-6) <J d2x + b2y + c2z = k2,
If these equations are multiplied, respectively, by
and the resulting equations are added, the sum is
(30-7)
= kib-2.Cz — ^i6
The coefficient of x in (30-7) can be denoted by the symbol
(30-8) D ^
0,2 62
+
This symbol is called a determinant of the third order. It is
also the determinant of the coefficients of the system (30-6).
Using the notation of (30-8), Eq. (30-7) can be written as
Dx s
61
b2
63
x =
k2 b2 c2
Similarly it can be shown that
bi GI
! b2 c2
y =
and
di b\ (
d2 62 <
^3 &3 <
^2 ^2 ^2 ,
!>!*!
<)2k2.
hkt
§30
SOLUTION OF EQUATIONS
105
If D 5^ 0, the unique solutions for xf y, and z can be obtained as
bi ki
(30-9) x =
0,2 ^2 C2
fc3 c3
D
z =
a* 62
,b,
In order to show that the values of x, y, and z, given in (30-9),
actually satisfy Eqs. (30-6), these values can be substituted in
the given equations.
If D = 0, the three equations (30-6) are either inconsistent or
dependent. A detailed analytic discussion of these cases will be
given in Sec. 35. Since the three equations of (30-6) are the
equations of three planes, a geometrical interpretation will now
be given.
If the three equations are inconsistent, the three planes are
all parallel, or two are parallel and are cut by the third plane in
two parallel lines. In either case, there is obviously no solution
for x, yj and z. If the equations are dependent, all three planes
intersect in the same line or all three planes coincide. In either
case there will be an infinite number of solutions for x, y, and z.
Example. For the system
Therefore,
3x- y - z = 2,
x - 2y - 3z = 0,
4x + y + 2z = 4.
D =
3 -1 -1
100
JL ~~~ £i """" O
2 -1
-1
0 -2
-3
4 1
2
2
2
2 *
3
-1
2
1
-2
0
ft
4
1
4
y =
2 -i
0 -3
4 2
= 2.
2,
PROBLEMS
1. Evaluate
1 2
2 -1
3 -1 -2
2 0 -3j
1 4 2|
-1 1 -2!
, and
4 -2 1
5 0-1
2 3-3
2. Find the solutions of the following systems of equations by using
determinants:
106 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §31
(a) 5s - 4y = 3,
2x + 3y = 7;
(&) 2z + 3y - 2* = 4,
x + y - z = 2,
3z - 57/ + 3s = 0;
(c) 3x - 2y = 7,
3?/ + 2z = 6,
2z + 3z = 1;
(d) 3x + 2y + 2z = 3,
z - 4y + 2z = 4,
2x + y + z = 2.
31. Determinants of the nth Order. Determinants of the
second and third orders were defined in the preceding section.
These are merely special cases of the definition of the determinants
of any order n. Instead of a symbol with 22 or 32 elements, the
determinant of the nth order is defined as the symbol, with n
rows and n columns,
#21
D =
#2n
#n,l #n2 * ' * #nn
which stands for the sum* of then! terms ( — l^a^ia^ * * ' #/cnn,
where &i, &2, • • • , kn are the numbers 1, 2, • • • , n in some
order. The integer k is defined as the number of inversions of
order of the subscripts fci, &2, • * • , kn from the normal order
1, 2, • • • , n, where a particular arrangement is said to have k
inversions of order if it is necessary to make k successive inter-
changes of adjacent elements f in order to make the arrangement
assume the normal order. There are nl terms since there are n!
permutations of the n first subscripts. Moreover, it is evident
that each term contains as a factor one and only one clement from
each row and one and only one element from each column.
* This sum is sometimes called the expansion of the determinant.
t It should be noted that it is not necessary to specify that the inter-
changes should be of adjacent elements, for it can be proved that, if any
particular arrangement can be obtained by k interchanges of adjacent
elements and also by'fc' interchanges of some other type, then k and kf are
always either both even or both odd. Hence, the sign of the term is inde-
pendent of the particular succession of interchanges.
§32 SOLUTION OF EQUATIONS
Example. Consider the third-order determinant
107
D =
l #12 #13
#21 #22 #23
#31 #32 #33
The six terms of the expansion are, apart from sign,
#11#22#33>
#21#32#13,
#11#32#23,
#31#12#23,
#2i#12#33,
#3l#22#13-
The first term, in which the first subscripts have the normal order, is
called the diagonal term, and its sign is positive. In the second term
the arrangement 132 requires the interchange of 2 and 3 to make it
assume the normal order; therefore, k — 1, and the term has a negative
sign. Similarly, the third term has a negative sign. The fourth term
will have a positive sign, for the arrangement 231 requires the inter-
change of 3 and 1 followed by the interchange of 2 and 1 in order to
assume the normal order. Similarly, it appears that the fifth term will
have a positive sign. In the sixth term, it is necessary to make three
interchanges (3 and 2, 3 and 1, and 2 and 1) in order to arrive at the
normal order; hence, this term will have a negative sign. As a result of
this investigation, it follows that
D = ttutt22#33 —
23 — #21#12#33 + #2i#,*2#13 + #31#12#23 ~ #31#22#13
It is evident that if k is equal to zero or an even number the
term will have a positive sign, whereas if k is odd the term will
be negative.
PROBLEM
Find the signs of the six terms involving #u in the expansion of the
determinant
#11 #12 #13 #14
#21 #22 #23 #24
#31 #32 #33 #34
#41 #12 #43 #44
32. Properties of Determinants. 1. The value of a determi-
nant is not changed if in the symbol the elements of corresponding
rows and columns are interchanged.
If
D s
«21
a2n
ani <
108 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §32
then the determinant formed by interchanging the corresponding
rows and columns is
an
T\l ...
Any term ( — l)*afc la* 2 • • • ak n of D, where &i, fc2, • • • , kn are
the numbers 1, 2, • • • , n in some order, will correspond to a
term ( — 1)^1^2 * * ' #/cnn of D', for each determinant must
contain every possible term that is a product of one and only
one element from each row and each column. But the number
of inversions is the same for the term of D as it is for the term of
ZX, owing to the fact that the corresponding first subscripts are
the same. It follows that each term of D occurs also in £)', and
conversely each term of Df occurs also in D.
Example. If
then
D'
2 5
1 -1
-3 -2
2 1 -3
5 -1 -2
3 4 1
-66,
= -66.
2. An interchange of any two rows or of any two columns of a
determinant will merely change the sign of the determinant.
If D is the original determinant and D" is the determinant
having the ith and jth rows of D interchanged, then the expansion
of D" will have the first subscripts of each term the same as those
of the corresponding term of D, except that i and j will be inter-
changed. Since it requires one interchange to restore i and j to
their original order in each term, the sign of every term will be
changed. Thus, D" = -D.
Example. If
then
2 5
1 -1
-3 -2 1
2 5
-3 -2
i —1
= 66.
SOLUTION OF EQUATIONS
109
3. // any two rows or any two columns of a determinant are
identical, the value of the determinant is zero.
For, by property 2, if these two rows (or columns) were inter-
changed, the sign of D should be changed. But since these two
rows (or columns) are identical, D remains unchanged. There-
fore, D = — D, and hence D = 0.
Example. If
then
2 -1
3 4
-2 5 -2
D = 0.
4. // each element of any row or any column be multiplied by
m, the value of the determinant is multiplied by m.
This follows from the definition of the determinant. Since one
and only one element of any row or column occurs in each term,
each term will be multiplied by m and therefore the value of the
determinant is multiplied by m.
Example 1. If
5
and
2
1
-3
D =
-2
-66
2 5
i i
j. j.
-6 -4
which has each element of the last row twice the corresponding element
of the last row of D, then
D = -132 and D = 2D.
Example 2. If
6
9
-6
4
2
3
then
= 2
3
9
-6
4
1
i
= 2-3
1
3
-2
3 -1
5. From properties 3 and 4, it follows that the value of a deter-
minant is zero if any two rows or any two columns have corre-
sponding^ elements proportional.
110 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §33
6. The product of two determinants D and D', both of order n,
is the nth-order determinant D" which has as the element in its
ith row and jth column the sum
ainbn
which is formed by multiplying each element alk of the ith row of D
by the corresponding element bk} of the jth column of D' .
Thus, if
D =
an
and
£>' =
611 6
621 622
then
D • D' s D" =
a\\b\i + 012621 #11612 + #12622
+ 022621 #21612 H~ ^22622
Example. The product of the following determinants is easily found
by expanding the product determinant:
sin
X
COS X
1
sir
sec
X
tan x
1
CO
CSC
X
cot x
1
—
0
tan
X
+ sin
x - 1
cos
X
cot x
sec x esc x
— tan x — cot x
- 1 - 1 - 1
tan x — sin x — 1
0
sec x esc x — 2
— cos x cot #
sec x esc £ — 2
0
- 2 cos2 x(2 — sec a? esc x).
33. Minors. The method of evaluating a determinant by the
use of the definition of Sec. 31 is exceedingly tedious, especially
if n ^ 4. There are other schemes for this evaluation, and these
require the definition of the minors of a determinant. The
simplest of these schemes will be described and used here.
'If, in the determinant Z>, the ith row and the jth column be
suppressed, the resulting determinant AlJ (which is of order one
less than the order of D) is called the minor of the element al3,
which is in the ith row and jth column.
Example. If
ai2 ais
#21 #22 a23 024
i 032 flss
1 tt42 tt43 #44
§33 SOLUTION OF EQUATIONS 111
then
flll #12
A23 ~ &31 &32
fl41 «42
From the definition of a determinant, it is evident that o,,AtJ is
composed of all the terms of D which contain the element atj as a
factor, except for the possibility that all the signs may be reversed.
Then the expression ( — l)&1anAn is composed of all the terms of
D containing an as a factor; ( — l)t2a2iA2i is composed of all the
terms containing a2i as a factor; (— l)*3o.3iA3i is composed of all
the terms containing a3i as a factor; etc. But D is composed of all
the terms containing an, a2i, a3i, • • • , ani as a factor, and so,
D = ( — l)*iaiiAn + ( — l)*2a2iA2i + • • • + ( — l)*na»iA»i.
It can be proved* that fci = 1 4- 1. fc2 = 2 + 1, &3 = 3 + 1,
• • • , kn = n + 1, so that
D = aiiAn - a2iA2i + • • • + ( — l)n+1aniAni.
In the above development for D the elements an, a2i, • • • , ani
are the elements of the first column of D. Similarly, the value
of D can be formed by taking the elements of any other column
or of any row.
Using the ith column gives
D = ( — IJ^ai.Au + (-l)fc2a2lA2t + • • • + (-l)fc*amAn;,
where ki = i + 1, fc2 = i + 2, • • • , kn = i + n. Similarly,
using the ith row gives
D = (-!'
where ki = i + 1, fc2 = i + 2, • • • , kn = i + n. It may be
observed that each kr is equal to the sum of the subscripts of its
at? and is thus equal to the sum of the number of the row and the
number of the column in which this element occurs. This
development is known as the expansion by minors, or the simple
Laplace development.
Since the term cofactor is frequently used in applications of
this type of development, it will be defined here. The cofactor
Ca of an element al3 is defined as the signed minor, that is,
* DICKSON, L. E., First Course in Theory of Equations, pp. 101-127;
FINE, H. B., College Algebra, pp. 492-519.
112 MATHEMATICS FOR ENGINEERS AND PHYSICISTS {33
Thus, the expression for D can be written as
n n
0=£(-l)'-H«.^ =
or as
n n
D = 2 ^A/ = 2 aijCij-
3406
0521
0340
= 3
521
340
1271
271
= 3[- 3
-0
406
340
2 7 1
+ 0
406
5 2 1
2 7 1
5 1
2 1
-9(2 - 7) + 12(5 - 2) - 4(0
13.
- 1
2 1
406
5 2 1
340
-0
5 1
3 0
5 2
3 4
4) - 6(20 - 6)
Here, the first expansion is made by using the elements of the first
column, for it contains two zeros (the third row is an equally good
choice). The expansion of the first third-order determinant is made by
using the elements of the second row, but the third column could be
used to equal advantage. In the expansion of the last third-order
determinant the first row was chosen, but the third row and the second
and third columns provide equally good choices.
The following theorem is given here because of its frequent use
in many fields of pure and applied mathematics :
n
THEOREM. The sum 2 a>%3Ck3 is zero, if k ^ i.
j-i
Each term of this sum is formed by taking the product of the
cofactor of an element of the fcth row by the corresponding
element of the ith row. This is the expansion of a determinant
whose t'th and fcth rows are identical and whose value is accord-
ingly zero. Similarly, it follows that S
= 0, if k ^ j.
Exampk. Let
Then,
D
3-1 2
1 2 -1
4 -3 -2
•7, Ciz - -2,
C
13
-11
§33 SOLUTION OF EQUATIONS 113
and the sum
3
2) aSjCi, = -28 + 6 + 22 = 0.
Similarly,
3
2) 02,Ci, = -7 - 4 + 11 = 0.
By using the theory of determinants, the solution of a system
of n non-homogeneous linear equations in n unknowns can be
obtained. The rule for effecting the solution will be stated but
not proved.* The proof for the cases when n = 2 and n = 3 has
already been given in Sec. 30.
Cramer's Rule. Let
(33-1)
== fo2
1 t^nn^n
be a system of n equations in the unknowns x\9
such that the determinant
D =
• • /)•
y An
of the coefficients is not zero. The system (33-1) has a unique
solution given by
D,
x -
Xn ""
where Z3t is the determinant formed by replacing the elements
«ii, a2l, ast> • • • , a«* of the tth column of D by fci, fc2, fc3, • • * ,
fcn, respectively.
Example. Solve, by Cramer's rule, the system
3z + y + 2z = 3,
2z - 3y - z= -3,
a? + 2t/ + 2 = 4.
* DICKSON, L. E., First Course in Theory of Equations, pp. 114-115.
114 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §34
Here
and
3 1
2
D = 2 -3 -
1 = 8
1 2
1
3 1 2
332
-3 -3 -1
2 -3 -1
4 2 1
8 ,
1 4 1
8
~~ o ~ 1> 2/
8
3 1 3
2 -3 -3
1 2 4
-8
z —
8
8 ~" '
PROBLEMS
1. Evaluate
1
-2
0
3
2
-1
3
0
—
1
4
1
2
1
4
0
2
2
0 -
-1 -3'
1
1
3
5
3
1
0
1
-1
0
2
0
2. Solve, by Cramer's rule,
the
systems
(a) x -{- 2y + 3z =
= 3,
(6) 2x +
2x -
y +
2 = 6, 3# —
3o; +
y —
z -
= 4.
# -
<<0
i
2y
= 1,
(d) 1
2x +
2x -
y -
22
= 3,
«
)X +
/>• i
2/ +
32
= 2.
a;
— 2
, and
16
132-1
04—32
-3 1 0 I1
120-4
- 3z = 2,
- 2z = 1,
x - y + z = — 1.
+ y + 3z + w = -2,
— 2 — W — 1,
34. Matrices and Linear Dependence. In order to discuss
the systems arising in the succeeding sections, it is convenient to
give a short introduction to the theory of matrices.*
An m X n matrix is defined as a system of mn quantities al}
arranged in a rectangular array of m rows and n columns. If
m = n, the array is called a square matrix of order n. The
quantities alj are called the elements of the matrix. Thus,
(34-1) A
an
a2n
or
an
a2n
#wl Clm2 * * ' a
* For detailed treatment see M. Bocher, Introduction to Higher Algebra,
pp. 20-53; L. E, Dickson, Modern Algebraic Theories, pp. 3&~63.
§34 SOLUTION OF EQUATIONS . . 115
where double bars or parentheses are used to enclose the array of
elements. If the order of the elements in (34-1) is changed or if
any element is changed, a different matrix results, Any two
matrices A and B are said to be equal if and only if every element,.
of A is equal to the corresponding element of J5, that is, if
al3 = btj for every i and j.
If the matrix is square, it is possible to form from the elements
of the matrix a determinant whose elements have the same
arrangement as those of the matrix. The determinant is called
the determinant of the matrix. From any matrix, other matrices
can be obtained by striking out any number of rows and columns.
Certain of these matrices will be square matrices, and the
determinants of these matrices arc called the determinants of the
matrix. For an m X n matrix, there are square matrices of
orders 1, 2, • • • , p, where p is equal to the smaller of the
numbers m and n.
Example. The 2X3 matrix
^_ /aH
\a-2l
contains the first-order square matrices (an), (at2), (a2s), etc., obtained
by striking out any two columns and any one row. It also contains
the second-order square matrices
obtained by striking out any column of A.
In many applications, it is useful to employ the notion of the
rank of a matrix A. This is defined in terms of the determinants
of A. A matrix A is said to be of rank r if there exists at least
one r-rowed determinant of A that is not zero, whereas all deter-
minants of A of order higher than r are zero.*
Example. If
/ 1 0 1 3\
EE( 2 1 0-21
\-l -1 1 5/
* In case an m X n matrix contains no determinants of order higher than r,
obviously r is the smaller of the numbers m and n, and the matrix is said
to be of rank r.
M6 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
the third-order determinants are
§34
I 0 1
1 0 3
1 1 3
2 1 0
= 0,
2 1 T-2
= 0,
20-2
= 0,
-1 -1 1
-1 -1 5
-1 1 5
0 1 3
1 0 -2
1 1 K
J. A c)
0.
Since
1
5*0,
there is at least one second-order determinant different from zero,
whereas all third-order determinants of A are zero. Therefore, the
rank of A is 2.
It should be observed that a matrix is said to have rank zero
if all of its elements are zero.
The notion of linear dependence is of importance in connection
with the study of systems of linear equations, and it will be con-
sidered next.
A set of m, m > 2, quantities /i, /2, /a, • • * , fm (which may be
constants or functions of any number of variables) is said to be
linearly dependent if there exist m constants ci, c2, • • • , cm, which
are not all zero} such that
(34-2) ci/i + c2/2 + • • • + cmfm s 0.
If no such constants exist, the quantities /t are said to be linearly
independent.
Example. If the /, are the polynomials
fi(x, y, z) = 2z2 - 3xy + 4z,
f2(x, y, z) 55 a?* + 2xy - 3z,
MX, y, z) ss 4z2 + xy- 2z,
and if the constants are chosen as ct = 1, 02 = 2, c3 = — 1, then
Cifi + c2/a + c3/3 ss 0.
Therefore, these three polynomials are linearly dependent.
It is evident that, whenever the set of quantities is linearly
dependent, at least one of the /t can be expressed as a linear
combination of the others. Thus, from (34-2), if ci ^ 0, then
§35 SOLUTION OF EQUATIONS 117
where
a2 = -, a3 = -> etc.
The definition of linear dependence requires the existence of at
least one constant cl ^ 0, and therefore the solution for /» is
assured.
Obviously, in most cases it would be extremely difficult to
apply the definition in order to establish the linear dependence
(or independence) of a given set of quantities. In case the
quantities ft are linear functions of n variables, there is a simple
test which will be stated without proof.*
THEOREM. The m linear functions
f^ z= al\x\ + alzXz + ' ' ' + QinXn, (i = 1, 2, • • • , m),
are linearly dependent if and only if the matrix of the coefficients
is of rank r < m. Moreover, there are exactly r of the fl that form
a linearly independent set.
If m > n, obviously r < m, and it follows that any set of m
linear functions in less than m unknowns must be linearly
dependent.
The fact that the polynomials
/i = 2x — 3y + 42,
/2 = x + 2y - 82,
/3 = 4s + y - 22,
are linearly dependent can be determined by observing that the matrix
of the coefficients,
/2 -3
-3 4\
2 -3I»
1 -2/
is of rank 2.
35. Consistent and Inconsistent Systems of Equations. A set
of equations that have at least one common solution is said to
be a consistent set of equations. A set for which there exists no
common solution is called an inconsistent set.
The question of consistency is frequently of practical impor-
tance. For example, in setting up problems in electrical net-
works, there are often more conditions than there are variables.
* DICKSON, L. E., Modern Algebraic Theories, pp. 55-60; BOCHER, M ,
Introduction to Higher Algebra, pp. 34-38.
118 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §35
This leads to a system in which there are more equations than
there are unknowns. It is important to have a method for
testing whether all the conditions can be satisfied simultaneously.
THEOREM 1. Consider a system of m linear equations in n
unknowns,
/0-. 1 \ a%lXl ~T~ #22#2 ~T~ •••-{- a%nXn == ^2»
(35-1)
;
_ _j_ (J,mnXn == K"mj
where at least one kt j^ 0. If the matrix of the coefficients is of
rank r, Eqs. (35-1) are consistent provided that the rank of
is also r.
The matrix K is called the augmented matrix. The proof of
this theorem will be found in any standard work on higher
algebra.
Example 1. Consider the system
2x + 3y = 1,
x - 2y = 4,
4x — y = 9.
Since
/2 3\
»(l -2)
\4 -I/
is of rank 2, the equations are consistent if
K
/2 3 1\
= (1 -2 4)
\4 -1 9/
is also of rank 2. This condition is satisfied; for the determinant of K
is zero, and there exists a second-order determinant of K that is dif-
ferent from zero.
Example 2. The system
2x + 3y = 1,
x - 2y = 4,
4s - y = 6
§35 SOLUTION OF EQUATIONS 119
is inconsistent, because
/2 3 1\
K = ( 1 -2 4 I
\4 -1 67
is of rank 3, whereas the matrix A is of rank 2.
In the case in which there are n equations in n unknowns, the
theorem on consistent equations shows that if the determinant of
A is zero, so thjt the rank of A is r < n, then the rank of K must
be r also, if the set of equations is to be consistent. If the rank of
K is greater than r, the set of equations is inconsistent. This
provides the analytic discussion that should accompany the
geometric discussion given forn = 3 in Sec. 30.
If the set of equations is consistent and the rank of A is r,
then it can be shown that n — r of the unknowns can be given
arbitrary values, and the values of the remaining r unknowns are
determined uniquely in terms of those n — r arbitrary values.
These n — r unknowns cannot be chosen at random, for the
m X T matrix of the coefficients of the remaining r unknowns must
have rank r if these unknowns are to be uniquely determined.
Example 3. Solve the system
x - y + 2z = 3,
x + y - 2z = 1,
x + 3y - 6z = -1.
Since A and K are botli of rank 2, the equations are consistent. If
either y or z is chosen arbitrarily the matrix of the coefficients of the
remaining variables will have rank 2. If z = k, the equations to
be solved are
x - y = 3 - 2k,
x + y = 1 + 2/b,
x + 3y — — 1 + 6k.
Solving the first two for x and y gives x = 2 and y = 2k — 1. These
values are seen to satisfy the third equation. Therefore, the solutions
x = 2, y — 2k — 1, z — k satisfy the original system for all values of k.
The preceding discussion has dealt with non-homogeneous
linear equations. In case the fct are all zero, the system becomes
the set of homogeneous equations
^11X1 + ai2#2 + * * ' + &ln£n ^ 0,
(35-2) a*lXl a™X* . a*nXn ""' '
. . . . ,
+ _!_ • 1 n ~ C\
(Z«t2«^*2 "i * * * "T~ ^*i»n«*'f> """" U«
120 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §35
Obviously, Zi = £2 = • • • = xn = 0 is a solution of (35-2).
It may happen that there are other solutions. If a\, a%, - - - , an
is a solution of (35-2), it is evident that fcai, fca2, • • • , kan, where
k is an arbitrary constant, will be a solution, also. The condition
for solutions different from the x\ = x% = • • - = xn = 0 solu-
tion will be stated without proof.
THEOREM 2. The system (35-2) will have a solution different
from the solution x\ = x% = • • • = xn = 0, if the rank of the
matrix of the coefficients is less than n.
It follows that if the number of equations is less than the
number of unknowns, that is, if m < n, there are always solutions
other than the obvious zero solution. If m = n, there exist
other solutions if the determinant of the square matrix of the
coefficients is zero. As in the case of the non-homogeneous
system, if the m X n matrix of the coefficients is of rank r, then
n — r of the unknowns can be specified arbitrarily and the
remaining r unknowns will be uniquely determined, provided that
the rank of the matrix of the remaining unknowns is r.
Example 4. Consider the system
2x - y + 3z = 0,
x + 2y — z = 0,
3x + 4y + z — 0.
Here
2-1 3
\A\ = 1 2 -1 = 10.
|3 4 1
Therefore, x - 0, y = 0, z = 0 is the only solution.
Example 5. Consider
3x - 2y = 0,
x + 4y = 0,
2x - y = 0,
for which the matrix of the coefficients is of rank 2. Since the number
of unknowns is 2, x = 0, y = 0 is the only solution.
Example 6. Consider
2x - y + 3z = 0,
x + 3y - 2z « 0,
5x 41 V + 4* - 0.
§36 SOLUTION OF EQUATIONS 121
Here,
A s
which is of rank 2. Since the number of unknowns is 3, the system
has solutions other than x = 0, y = 0, z = 0. Let z = k, and solve
any two of the equations for x and y. If the first two are chosen,
x = — k and y = k. By substitution, it is easily verified that x = — k,
y = k, z = k satisfies all four equations for any choice of k.
Example 7. Consider
2x - 4y + 2 = 0,
3x + y - 2z = 0.
For this system,
A /2 ~4 ^
A = (3 1 -2J'
which is of rank 2 Since the number of unknowns is greater than the
number of equations, there exist other solutions. Let z = k, and solve
the two equations for x and y. There results x — %k and y = %k.
Thus, x = %k, y = J^&, 2 = k is a solution for any choice of k.
Example 8. Consider
x - y + 2 = 0,
2x + 3y + 2 = 0,
3s + 2?/ + 2z = 0.
Here,
\A\ =
2 3 1
322
Since the determinant of A is zero, there are solutions different from
x = 0, y = 0, z = 0. Let z = fc, and solve any two of the equations.
If the first two are chosen, x = — ££&, ?/ == K&> z — k. It is verifiable
by substitution that these values satisfy all three equations, whatever
be the choice of k.
PROBLEMS
1. Investigate the following systems and find solutions whenever the
systems are consistent:
(a) x - 2y = 3, (6) 2x + y - z = 1,
2x + y = 1, x - 2y + z = 3,
3x — y = 4. 4x — 3y + z = 5.
(c) 3x + 2y = 4, (d) 2x - y + 3z = 4,
a? — 3y = 1, s + If — 3« = -1,
2z + 6y = -1. 5s - y + 3z = 7.
122 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §35
2. Investigate for consistency, and obtain non-zero solutions when
they exist.
(a) x + 3y - 2z = 0, (b) x - 2y = 0,
2x - y + z = 0. 3x + y = 0,
2z - y = 0.
(c) 3x -2y + 2 = 0, (d) 2o; - 4y + 82; = 0,
a; + 2y - 2,3 = 0, x + 2?y - 2z = 0,
2x — y + 2z = Q. 3x - 2?/ + 2 = 0.
(e) 4a; - 2y + 2 = 0, (/) x + 2?y + 2z = 0,
2x - y + 3z = 0, 3x - y + z = 0,
2x - y - 2z = Q, 2x + 3y + 2z = 0,
Qx — 3y + 42 = 0. x + 4i/ - 2z = 0.
CHAPTER IV
PARTIAL DIFFERENTIATION
36. Functions of Several Variables. Most of the functions
considered in the preceding chapters depended on a single
independent variable. This chapter is devoted to a study of
functions depending on more than one independent variable.
A simple example of a function of two independent variables
x and y is
z = xy,
which can be thought to represent the area of a rectangle whose
sides are x and y. Again, the volume v of a rectangular parallel-
epiped whose edges are x, y, and z, namely,
v = xyz,
is an example of a function of three independent variables x,
y, and z. A function u of n independent variables x\, x^ • • • ,
xn can be denoted by
U = f(Xi, X2, ' ' • , Xn).
A real function of a single
independent variable x, say
V — /(#)> can be represented
graphically by a curve in the
xy-pl&ne. Analogously, a real
function z = f(x, y), of two
independent variables x and
Fio. 27.
y, can be thought to represent a surface in the three-dimen-
sional space referred to a set of coordinats axes x, y, z (Fig.
27). However, one must not become too much dependent
on geometric interpretations, for such interpretations may
prove to be of more hindrance than help. For instance,
the function v = xyz, representing the volume of a rectangular
parallelepiped, depends on three independent variables x, y,
123
124 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §36
and z and hence cannot be conveniently represented geometrically
in a space of three dimensions.
Corresponding to the definition of continuity of a function
of a single independent variable x (see Sec. 7), it will be said that
a function z = f(x, y) is continuous at the point (XQ, yo) provided
that a small change in the values of x and y produces a small
change in the value of z. More precisely, if the value of the
function z = /(#, y) at the point (XQ, yd) is z0, then the continuity
of the function at the point (x0, yo) means that*
(36-1) lim f(x, y) = f(xQ) yo) = z0.
In writing the left-hand member of (36-1), it is assumed that the
limit is independent of the mode of approach of (x, y) to (XQ, y0).
The statement embodied in (36-1) is another way of saying
that
f(x> y} ~ f(xv> yo) + €,
where lim e = 0; that is, if the function /(x, y) is continuous at
x—>xo
(XQ, yo), then its value in the neighborhood of the point (XQ, yo) can
be made to differ from the value at the point (XQ, yo) by as little as
desired.
If a function is continuous at all points of some region R
in the :n/-plane, then it is said to be continuous in the region R.
The definition of continuity of a function of more than two
independent variables is similar. Thus, the continuity of the
function u = f(x, y, z) at the point (x0, yo, z0) means that
lim f(x, y, z) = /(zo, y0, z0),
z— >ro
y— *i/o
Z—+ZO
independently of the way in which (x, y, z) approaches (x0, y0, z0).
PROBLEM
Describe the surfaces represented by the following equations:
(a) x + 2y = 3, (6) x - y + z = 1, (c) x - 2, (d) z = y,
(e) 2x-3y + 7z = 1, (/) x2 - ?/ = 0, (g) y* + z* = 25,
(A) i/2 = 2x, (i) z2 + ?/ - 10s = 0, (/) z2 + 2/2 + z2 = 1,
(AO 3* + z2 = y, (0 z2 + 27/2 + z = 0, (m) z2 + 7/2 = z2,
* For details, see I. S. Sokolnikoff, Advanced Calculus, Chap. III.
§37
PARTIAL DIFFERENTIATION
125
y2
T
(r)
37. Partial Derivatives. The analytical definition of the
derivative of a fuhctiori y = f(x), of a single variable x, is
. Ihn = Km
This derivative can be interpreted geometrically as the slope of
the curve represented by the equation y = f(x) (Fig. 28).
z^
It is natural to extend the definition of the derivative to
functions of several variables in the following way: Consider
the function z = f(x, y) of two independent variables x and y.
If y is held fast, z becomes a function of the single variable x
and its derivative with respect to x can be computed in the
usual way. Let AZ* denote the increment in the function
z = /(x, y) when y is kept fixed and x is changed by an amount
Ax; that is,
Az* = f(x + Ax, y) - /(x, y).
Then,
lim * _ lim -
Ax-»0 AX Az-»0 AX
is called the partial derivative of z with respect to x and is denoted
by the symbol dz/dx, or zx, or fx.
126 MAfKEMATICS FOR ENGINEERS AND PHYSICISTS §37
Similarly, the partial derivative of z with respect to y is defined
by
In general, if w = /(xi, 0:2, • • • , xn) is a function of n inde-
pendent variables x\y x2, • • * , xn, then du/dxt denotes the
derivative of u with respect to xt when the remaining variables
are treated as constants. Thus, if
z = x3 + x2y + 2/3,
then
g = 3x2 + 2xy and g = x2 + 3</2.
Also, if u = sin (ax + by + cz), then
— = a cos (ax + by + cz) (both t/ and 2 held constant) ;
CfX
— = 6 cos (ax + by + cz) (both x and 2 held constant) ;
— = c cos (ax + by + cz) (both x and ?/ held constant).
dz
In the case of z = /(x, ?/), it is easy to provide a simple geo-
metric interpretation of partial derivatives (Fig. 27). The equa-
tion z = /(x, T/) is the equation of a surface; and if x is given the
fixed value x0, 2 = /(x0, y) is the equation of the curve AB on
the surface, formed by the intersection of the surface and the
plane x = XQ. Then dz/dy gives the value of the slope at any
point of AB. Similarly, if y is given the constant value y0,
then z — /(x, T/O) is the equation of the curve CD on the surface,
and dz/dx gives the slope at any point of CD.
PROBLEMS
1. Find dz/dx and dz/dy for each of the following functions:
(a) z = y/x\ (b) z = x3?/ + tan-1 (y/x)\ (c) z = sin xy + x;
(d) z = e* log y; (e) z == x2y + sin"1 x.
2. Find du/dx, du/dy, and du/dz for each of the following functions:
(a) u = x2y + yz - zz2; (6) u = xyz + log xy;
(c) it = z sin-1 (x/y); (d) u = (x2 + y2 + z2)^;
(e) it = (x* + 2/2 + s2)-^.
§38 PARTIAL DIFFERENTIATION 127
38. Total Differential. In the case of a function of one
variable, y = /(a:), the derivative of y with respect to x is defined
as
so that Ay/ Ax = /'(a;) + €, where lim e = 0. Therefore,
Ax— »0
/(x + Ax) - /(x) ss Ay = /'(x) Ax + e Ax,
where e is an infinitesimal which vanishes with Ax. Then,
is defined as the differential dy.
For the independent variable x, the terms " increment" and
"differential" are synonymous (that is, Ax = dx). However,
it should be noted that the differential dy (of the dependent
variable y) and the increment Ay differ by an amount c Ax (see
Fig. 28).
The differential of a function of several independent variables
is defined similarly. Let z = /(x, y), and let x and y acquire the
respective increments Ax and Ay. Then,
Az = /(x + Ax, y + Ay) - /(x, y).
If 2 = /(x, y) is a continuous function, then, as Ax and Ay
approach zero in any manner, Az also approaches zero as a limit.
It will be assumed here that /(x, y) is continuous and that
df/dx and df/dy are also continuous.
The expression for Az can be put in a more useful form by
adding and subtracting the term/(x, y + Ay). Then,
Az = /(x + Ax, y + Ay) - /(x, y + Ay) + /(x, y + Ay) - /(x, y).
But
lim /<>* + Ax, y + Ay) - /(x, y + Ay) = df(x, y + Ay)
AZ-+O Ax dx
so that
/(x + Ax, v + Ay) - f(x, y +
where lim ei = 0. Moreover,
Ay-,0
128 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §38
since the derivative is continuous. Therefore,
df(x, y + Ay) = a/Qc, y} ^
dx dx 2'
where lim €2 = 0.
Af/~ -»0
In like manner,
f(x, y + Ay) - /(re, y) =
where lim e' = 0. It follows that
e A*
in which € = ei + c2.
The expression
is defined as the total differential of z and denoted by cfc. In
general, if w = /Or i, #2, • • • , #n), the total differential is given by
(38-1) du = -^~dxl + -j¥-dx2 + ' ' ' + ^ dxn.
oXi 0X2 oXn
The expression for the total differential is called the principal part
of the increment Aw, and is a close approximation to Au for
sufficiently small values of dx\, dx%, • • • , and dxn. As in the
case of a function of a single independent variable, the differential
of each independent variable is identical with the increment of
that variable, but the differential of the dependent variable
differs from the increment.
If all of the variables except one, say xt, arc considered as
constants, the resulting differential is called the partial differ-
ential and is denoted by
, df .
•—-
The partial differential expresses, approximately, the change
in u due to a change A#t = dxl in the independent variable xl.
On the other hand, the total differential du expresses, approxi-
mately, the change in u due to changes dx\, dx^ • • • , dxn in all
§38 PARTIAL DIFFERENTIATION 129
the independent variables Xi, x%, • • • , xn. It may be noted
that the total differential is equal to the sum of the partial
differentials. Physically, this corresponds to the principle of
superposition of effects. When a number of changes are taking
place simultaneously in any system, each one proceeds as if it
were independent of the others and the total change is the sum
of the effects due to the independent changes.
Example 1. A metal box without a top has inside dimensions
6 X 4 X 2 ft. If the metal is 0.1 ft. thick, find the actual volume of
the metal used and compare it with the approximate volume found
by using the differential.
The actual volume is AF, where
AF = 6.2 X 4.2 X 2.1 - 6 X 4 X 2 = 54.684 - 48 - 6.684 cu. ft.
Since F = xyz, where x = 6, y = 4, z = 2,
dV = yz dx + xz dy + xy dz
= 8(0.2) + 12(0.2) + 24(0.1) - 6.4 cu. ft.
Example 2. Two sides of a triangular piece of
land (Fig. 29) are measured as 100 ft. and 125 ft., FIG. 29.
and the included angle is measured as 60°. If the
possible errors are 0.2 ft. in measuring the sides and 1° in measuring
the angle, what is the approximate error in the area?
Since A = %xy sin a,
dA = %(y sin a dx + x sin a dy + xy cos a da),
and the approximate error is therefore
dA ~ \ t
125 r (0-2) + 10° r
+ 100(125) ~ —Q = 74.0 sq. ft.
PROBLEMS
1. A closed cylindrical tank is 4 ft. high and 2 ft. in diameter (inside
dimensions). What is the approximate amount of metal in the wall
and the ends of the tank if they are 0.2 in. thick?
2. The angle of elevation of the top of a tower is found to be 30°, with
a possible error of 0.5°. The distance to the base of the tower is found
to be 1000 ft., with a possible error of 0.1 ft. What is the possible error
in the height of the tower as computed from these measurements?
130 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §39
3. What is the possible error in the length of the hypotenuse of a
right triangle if the legs are found to be 11.5 ft. and 7.8 ft., with a
possible error of 0.1 ft. in each measurement?
4. The constant C in Boyle 's law pv — C is calculated from the
measurements of p and v. If p is found to be 5000 Ib. per square foot
with a possible error of 1 per cent and v is found to be 15 cu. ft. with a
possible error of 2 per cent, find the approximate possible error in C
computed from these measurements.
5. The volume v, pressure p, and absolute temperature T of a perfect
gas are connected by the formula pv = RT, where R is a constant. If
T = 500°, p = 4000 Ib. per square foot, and v = 15.2 cu. ft., find the
approximate change in p when T changes to 503° and v to 15.25 cu. ft.
6. In estimating the cost of a pile of bricks measured as 6 X 50 X 4
ft., the tape is stretched 1 per cent beyond the estimated length. If
the count is 12 bricks to 1 cu. ft. and bricks cost $8 per thousand, find
the error in cost.
A
7. In determining specific gravity by the formula s = , _ ^
where A is the weight in air and W is the weight in water, A can be read
within 0.01 Ib. and W within 0.02 Ib. Find approximately the maxi-
mum error in s if the readings are A = 1.1 Ib. and W — 0.6 Ib. Find
the maximum relative error As/s.
8. The equation of a perfect gas is pv — RT. At a certain instant
d given amount of gas has a volume of 16 cu. ft. and is under a pressure
of 36 Ib. per square inch. Assuming R = 10.71, find the temperature
T. If the volume is increasing at the rate of H cu. ft. per second and
the pressure is decreasing at the rate % Ib. per square inch per second,
find the rate at which the temperature is changing.
9. The period of a simple pendulum with small oscillations is
r = 2TV^
If T is computed using I = 8 ft. and g — 32 ft. per second per second,
find the approximate error in T if the true values are I — 8.05 ft. and
g = 32.01 ft. per second per second. Find also the percentage error.
10. The diameter and altitude of a can in the shape of a right circular
cylinder are measured as 4 in. and 6 in., respectively. The possible
error in each measurement is 0.1 in. Find approximately the maxi-
mum possible error in the values computed for the volume and the
lateral surface.
39. Total Derivatives. Thus far, it has been assumed that x
and y were independent variables. It may be that x and y
§39 PARTIAL DIFFERENTIATION 131
are both functions of one independent variable t, so that z
becomes a function of this single independent variable. In
such a case, z may have a derivative with respect to t.
Let z = f(x, y), where x = <p(t) and y = $(()] these functions
are assumed to be differentiable. If t is given an increment A£,
then x, y, and z will have corresponding increments Az, Ay,
and Az, which approach zero with At. As in the case when x and
y were independent variables,
Az = |£ Ax + j£ Ay + 61 Ax + c2 Ay.
Then,
= _ Arc Ay
A£ dx ~A* "*" dy A* Cl AJ "*" €2 A*
and
cfe
Moreover, from (39-1) it appears that
dz = fdx+fdy
dx dy
gives the expression for the differential in this case as well as
when x and y are independent variables.
The general case, in which
Z = f(Xi, X2, * • • , Xn)
with
can be treated similarly to show that
and
In case ^ = a: (39-1) becomes
cfe _ ^f < = t
dx ~ fa Itydx ~ Ite Itydx
This formula can be used to calculate the derivative of a func-
132 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §39
tion of x defined implicitly by the equation f(x, y) = 0. Let
z — /(x, y), so that
* = # + # &
dx dx dy dx
Since z = f(x, y) = 0, it follows that dz/dx = 0 and
dy _
provided that d/,% ^ 0.
As an example, let
x*y + y*x - 1 = 0
define y as an implicit function of x. Then,
& a:2 + 2xy
for all values of x and y for which the denominator does not
vanish.
It was noted that the total differential of a function
z = f(xi, x2, • • • , Xn), where xt = <pt(£),
is given by
It will be proved next that the same formula can be applied to
calculate the differential even when the variables x% are functions
of several independent variables ti, t^ • • * , tm. Thus, consider
z = f(xi, X*, • - • , xn), where x% = <pt(ti, t2, • • • , tm).
In order to find the partial derivative of f(xi, x^ • • • , xn) with
respect to one of the variables, say fc, the remaining variables are
held fixed so that f(xi, x%, • • • , xn) becomes a function of the
single variable tk. Then,
(39-2) >
oj oXi . oj dx% , , oj uXn
df dxi , df dx% . . df dxn
^L ^1 _i_ ^L dx* 4. . . . J_ ^/ ^n
dxi dtm dx% dtm dxn dtm
§39 PARTIAL DIFFERENTIATION 133
If the first equation of (39-2) is multiplied by dti, the second
by dt%, etc., and the resulting equations are added, there results
or
(39-3) rf/ = ^^ + ^^+---+
This establishes the validity of the formula (38-1) in all cases
where the first partial derivatives, are continuous functions,
irrespective of whether the independent variables are Xit x%,
• - • , xn or ti, h, • • • , tm.
An important special case of the formula (39-8) arises in
aerodynamics and other branches of applied^ mathematics.
Consider a function u = f(x, y, z, t) of four variables x, y, z, and t.
The total differential of u is
(39-4) ^-J^ + I^ + IA+I*
du , , du , . du , . du j.
^*cdx + d-ydy+tedz+-dldt-
Let it be supposed that x, y, and z are not independent variables,
but functions of the variable t. In such a case, u will depend on
t explicitly, and also implicitly through x, y, and z. Dividing
both members of (39-4) by dt gives
fQo K\ du _ ^d^ .dudy .dudz du
( ' Tt "" toT* + ^"5? + a^di "*" df
On the other hand, if the variables x, y, and z are functions of t
and of some other set of independent variables r, s, • • • , one
must replace dx/dt, dy/dt, and dz/d£ in the right-hand member
of (39-5) by dx/dt, dy/dt, and dz/dt, respectively, and du/dt in
134 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §39
the left-hand member by du/dt. The partial derivative with
respect to t which appears in the left-hand member differs from
that appearing in the right-hand member, since the latter is com-
puted from u= f(x, y, z, t) by fixing the variables x, y, and z
and differentiating the resulting function with respect to t. In
order to indicate the distinction between the meanings of the
two partial derivatives with respect to ty one can write
Du _ du dx du dy dudz du
~~dt~~'dxdt dy'dt 'dzdt ~dt'
The fact that the total differential of a composite function
has the same form irrespective of whether the variables involved
are independent or not permits one to use the same formulas for
calculating differentials as those established for the functions
of a single variable. Thus,
d(u + v) = du + dv,
= v du + u dv,
•to
etc.
Example 1. If u = xy + yz + zx, and x = t, y = e~', and z — cos t,
du , , . dx , , , . dy , t , N dz
Tt = (y + ^Tt + (x + z^dt + (x + y) ~dt
= (e-< + cos 0(1) + (t + cos «)(-s~0 + (< + «"0(- sin 0
= e~* + cos t — te*' — e~( cos t — t sin t — e~* sin t.
This example illustrates the fact that this method of computing du/dt
is often shorter than the old method in which the values of x, y, and z
in terms of t are substituted in the expression for u before the derivative
is computed.
Example 2. If f(xy y) = x2 + i/2, where x — r cos <p and y — r sin <p,
then
df df dx df dv
dr = dxfr + dyfr = 2x cos ^ + 2y sin ^ = 2r a*1 * + 2r sinV = 2r,
^ = i^ + %l^= 2a;(^r sin ^ + 22/(r cos ^
= — 2r2 cos v? sin <p + 2r2 cos ^? sin <p = 0.
Also,
df = 2r dr or df =* 2xdx + 2y dy.
§39 PARTIAL DIFFERENTIATION 135
Example 3. Let z = e*v, where x = log (u + v) and y = tan"1 (u/v).
Then,
az az dx 1 , ay t;
-jp-r, = *e- - and -
Hence,
du ~ dx du dy du ~~ u + v v2 + u2
Similarly,
az _ yexv xexvu
~fo = u + V ~~ V2 + U2'
The same results can be obtained by noting that
dz — yexv dx + xexv dy.
But
, dx , dx , 1 . , 1 ,
dx — 3- du + 3- dv = — : — du -\ : — dv
du dv u + v u + v
and
dy — -5- du + ^ dv = -^-n — "9 ^ r— — r dv.
y du dv v2 + u2 v2 + u2
Hence,
. du + dv , v du — u dv
dz = yexv : h xexv — ;
y u + v v
= ( — i h -r~l — 2 J ^ "
\u -}- v v2 + 'W2/
But
,1 dz j i dz j
dz = -5- du + -~- dv:
du dv '
and since dz* and dv are independent differentials, equating the coeffi-
cients of du and dv in the two expressions for dz gives
a^ __ 7/e*" ze^z;
a^ u + t; * f 2 + u2
and
__
ay ~ tT+li "" v2 + u"2'
PROBLEMS
1. If u = z?/z and a; = a cos 0, |/ = a sin ^, 2 == kO, find du/dQ.
2. If ^ = a;2 — 7/2 and ?/ = r sin 6 and # = r cos 0, find aV^r and
3. If u = xy — ?/2J and a; = r + s, y = r — s, 2; = ^, find du/dr,
du/ds, and du/dt.
136 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §40
*?/
4. If z = e**, z = log v-w2 + y2, and z/ = tan"1 — i find dz/du and
6. If z = /(a; + M, y + *0» show that d2/dz = dz/du and
dz/dz>.
6. If u = x2y + y*z + z2x, verify that
du t du t du , . , xo
7. (a) Find du/dt, if u = e* sin yz and 2 = Z2, y = i — 1, z = I//.
(6) Find dw/dr and dw/d0, if ^ = x2 - 4?/ and x = r sec 0, y = r tan 0.
8. (a) Find dt//dz and du/dx, if ^ = x2 + y2 and t/ = tan x.
(b) Given V = /(a;; T/, 2;), where x = r cos 0, t/ = r sin 0, 2 = i. Com-
pute dF/dr, aF/a0; 57/5^ in terms of dV/dx, dV/dy, and dV/dz.
- ?y
9. If /is a function of u and v, where u = v#2 ~h 2/2 ^nd v = tan-1 ->
find olf/ax, a//^2/, and VW/dx)2 +
40. Euler's Formula. A function /(#i, a:2, • • • , xn) of n
variables xi, x^, • • - , xn is said to be homogeneous of degree m
if the function is multiplied by Xm when the arguments xi,
xz, - - • , xn are replaced by \Xi, Xx2, • • • , Xxn, respectively.
For example, f(x, y) = xz/-\/x2 + y2 is homogeneous of degree
1, because the substitution of \x for x and \y for y yields
Xz2/ V^M7?- Again, /(«, y) = i + log * """ log y is homo-
y %
geneous of degree — 1, whereas /(#, 2/, z) = z*/\/x* + y2 is
homogeneous of degree %.
There is an important theorem, due to Euler, concerning
homogeneous functions.
EULER'S THEOREM. If u = /(xi, x2, • • * , xn) is homogeneous
of degree m and has continuous9 first partial derivatives, then
The proof of the theorem follows at once upon substituting
x'i = Xxi, #2 = ^2, ' * * , xfn ==: Xxw.
Then, since /(rci, o:2, • * * , xn) is homogeneous of degree m,
f(x(, x'2, • • • , <) =
§41 PARTIAL DIFFERENTIATION 137
Differentiating with respect to X gives
Ij*1 + &,** + ' ' ' + &.Xn = m*n~lf(Xi> ** ' ' ' ' Xn)-
If X is set equal to 1, then x\ = x(, x* = x'2, - * • , xn = x'n and
the theorem follows.
PROBLEM
Verify Euler's theorem for each of the following functions:
(a) f(x, y, z) = x2y + xy2 + 2xyz\
(6) f(x, y) = vV - x* sin-1
N 1 , log a; - log y.
, y) = —t H -- - ;
(»)
41. Differentiation of Implicit Functions. It was noted in
Sec. 39 that the derivative of a function of x which is defined
implicitly by the equation f(x, y) = 0 could be calculated by
applying the expression for the total derivative. This section
contains a more detailed discussion of this method.
The equation f(x, y) = 0 may define either x or y as an implicit
function of the other. If the equation can be solved for y to
give y = <p(x), then the substitution of y = <p(x) in f(x, y} = 0
gives an identity. Hence, f(x, y) = 0 may be regarded as
a composite function of x, where x enters implicitly in y. If
so that
It will be observed that this discussion tacitly assumes that
f(Xj y) = 0 has a real solution for y for every value of x. If
(41-1) is applied formally to z2 + y* = 0, it is readily checked
138 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §41
that -—• = --- This result is absurd for real values of x and y,
dx y *"
inasmuch as the only real values of x and y that satisfy x1 + y2 = 0
are x = 0 and y = 0.
Example 1. Find dy/dx, if 3xsy* + x cos £/ = 0. Here,
cos y, -g~ = 6z3?/ — 2 sn
so that
rf^ __ 9o;2?/2 + cos y
__
dx ~~ ~~ tix*y — x sin y
The relation /(x, ?/, z) = 0 may define any one of the variables as
an implicit function of the other two. Let x and y be independent
variables. Then /(x, y, z) = 0 defines 3 as an implicit function
of x and i/, and
, dz , , dz ,
dz = ~ dx + — dy.
dx dy *
But
Therefore, by substitution,
This can be written as
Since dx and dt/ are independent differentials and the above rela-
tion holds for all values of dx and dy, it follows that
dx dz dx
and
dy dz dy
If df/dz 9* 0, these equations can be solved to give
Example 2. If a2 + 2y2 - Szz = 0, then, by (41-2),
SL m _ 2a; - 3g — « - 4^
da? "" — 3a; J dy ~~ -3x
§41
PARTIAL DIFFERENTIATION
139
(41-3)
Frequently, it is necessary to calculate the derivatives of a
function that is defined implicitly by a pair of simultaneous
equations
>, y, z) = o,
(*, y, 0 = o.
If each of these equations is solved for one of the variables, say
Zj to yield
z = F(x, y) and z = <£(#, y),
then one is led to consider the equation resulting from the
elimination of 2, namely,
F(x, y} - $(z, y} = 0.
This equation may be thought to define y as an implicit function
of Xj and one can apply the method discussed earlier in this
section to calculate dy/dx.
However, the elimination of one of the variables from the
simultaneous equations (41-3) may prove to be difficult, and it
is simpler to use the following procedure: The differentiation
of (41-3) gives
and
These equations can be solved for the ratios to give
dx \dy\dz —
dy dz
dy dz
dz dx
d<f> d<p
dz dx
dx dy
d<p d<p
dx 'dy
from which the derivatives can be written down at once.
Example 3. Let
f(x, y, z) s x2 + 7/2 + z* - a2 = 0
and
Then,
dx:dy: dz
2y 2z
-2y — 4z
2z 2x
-4z 2x
2x 2y\
2x -2y\
—4yz:12xz: —
140 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §41
Hence,
dy _
dz -*
etc.
dx — tyz dx —
Another important case arises from a consideration of a pair
of simultaneous equations
(41-4)
lf(x, y, u, v) = 0,
\ <p(x, y, u, v) = 0,
which may be thought to define u and v as implicit functions
of the variables x and y.
Differentiating (41-4) gives
' ~ dx dy ~ — " -
(41-5)
.
du
dv
But, since u and v are regarded as functions of x and y,
, du , . du ,
du = —dx + —dy
ox oy
and
, dv 7 . dv 7
ay = — aa; + — ai/.
da: dy y
Substituting for du and dv in (41-5) gives
»fdu , d/d^ , /3/ , a/aw
du dx
XT, ^ ) d* +
^ v w . d/ di^
+ a^a^ av^
/^ ^
\dx + Tu
dx
du
dv
Since the variables a; and y are independent, the coefficients
of dx and dt/ must vanish, and this leads to a set of four equations
for the determination of du/dXj du/dy, dv/dx, and dv/dy. Thus,
one obtains
dJL i
dx dv
d<p d<p
dx dv
du
~dx
df_df_
du dv
dtp d<p
du dv
§41
PARTIAL DIFFERENTIATION
141
and similar expressions for du/dy, dv/dx, and dv/dy. It is
assumed in the foregoing discussion that all the derivatives
involved are continuous and that
df_ tf
du dv
Example 4. If
then
du dv
+ yz + u* + v3 = 0,
1 + y - u* + v* = 0,
du
~dx
PROBLEMS
1. Obtain dy/dz, dw/dy, and dv/dy in Example 4, Sec. 41.
2. Compute dy/efo, if xz + ?/3 — 3xy = 1.
3. Find dy/dx if
a;2?/ — y2z +
4. Obtain du/dx and dv/dy, if
— a3 = 0.
ttev — zt/ + t; = 0,
vey — xv + u — 0.
5. If x =
to x gives
and ?/ =
then differentiation with respect
dx du dx dv
du dx dv dx
dy du dy dv
~~ dudx dv dx
from which du/dx and dv/dx can be computed. Consider the pair of
equations
x = u2 - v2,
y = uv,
and obtain du/dx, du/dy, dv/dx, and dv/dy.
6. Apply the method outlined in Prob. 5 to find du/dx, dv/dx,
du/dy, and dv/dy, if
j X = U + V,
\y = 3u + 2v;
(a} \,
( 2x = v2 — w2,
/ y — uv.
142 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §41
7. If x = r cos 6 and y = r sin 0, find dr/dz and d0/d#.
8. If w - uv and
f N i w2 4- r + x = 0,
(a) ) * - u - y = 0,
one can obtain dw/dx as follows: Differentiation of w with respect to
x gives -3- = u T- + 0 "r The values of du/dx and dt>/dz can be
calculated from (a) by the method of Prob. 5. Find the expressions
for dw/dx and dw/dy.
9. If z — uv and
u* -f- v2 - x - y = 0,
W* - 02 + 3S + # = 0,
find dz/dx.
10. If 2 = w2 + w2 and
2 = ^2 _ V29
y = w,
find dz/dx.
11. If 2 = n2 + f2 and
u = r cos 0,
y = r sin 0,
find dz/dr and dz/dO.
12. If r = (x2 + /y2)'2' and 0 = tan"1 -, find dr/dx and dO/'dx.
x
13. (a) Find du/dx, if x sec y + z3?y2 - 0.
(b) Find cte/da; and dz/dy, if z3?/ — sin 2 + 23 = 0.
14. Let u = x + y + z = 0 and «; == s2 + 7/2 + 22 - a2 = 0. lir <l
dx \dy\dz.
15. Find du/dx, dv/dx, du/dy, and dv/dy, if
16. Find dw/dx and dw/dy, ii w = u/v and
17. Show that ||f = 1 and ||f | = -1, if f(x,y,z) = 0.
Note that, in general, dz/dx and dz/dz are not reciprocals.
18. Find du/dx, if
u2 - v2 - x3 + 3y = 0,
-w + t; - y2 - 2x = 0.
19. Prove that
§42
PARTIAL DIFFERENTIATION
143
if F(x, y, u, v) = 0 and G(x, y, u, v) = 0.
20. Show that (g)1 + 1/r* (|)' - (|)' + (£)', if .
and y = r sin 0.
r cos
42. Directional Derivatives. The relation expressed in (39-1)
has an important special case when x and y are functions of the
distance s along some curve C, which goes through the point
(x, y). The curve C may be thought to be represented by a pair
of parametric equations
x = x'(s),
2/ = 2/0),
where x and y are assumed to possess continuous derivatives with
respect to the arc parameter s.
Let P (Fig. 30) be any point of
the curve C at which /(x, y) is
defined and has partial deriva-
tives df/dx and df/dy. Let
Q(x + Ax, y + Ay)
be ^a point close to P on this
curve. If As is the length of the
arc PQ and A/ is the change in / due to the increments Ax and
Ay, then
df r Af
~- = hm -~
as AS-»O As
gives the rate of change of/ along C at the point (x, y).
FIG
But
^ = y ^ a/ ^
ds ^x rfs ^y ^5 '
and
dx
-r-
as
,. Ax
hm ~~ = cos a.
A5->o As
dy ,. Ay
-r = hm -— = sin a.
ds A*-»O As
Therefore,
(42-1)
df df . df .
-r- = -— COS a + T~ Sin a,
as dx dy
and it is evident that df/ds depends on the direction of the curve.
For this reason, df/ds is called the directional derivative. It
represents the rate of change of / in the direction of the tangent to
144 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §42
the particular curve chosen for the point (x, y). If a = 0,
#= #,
ds dx
which is the rate of change of / in the direction of the x-axis. If
a = r/2,
df^V,
ds dy
which is the rate of change of / in the direction of the y-axis.
Let z = f(Xj y), which can be inter-
preted as the equation of a surface, be
represented by drawing the contour
lines on the zy-plane for various values
of z. Let C (Fig. 31) be the curve in
the #?/-plane corresponding to the value
2 = 7, and let C + AC be the neighbor-
ing contour line for z = 7 + &y.
+x Then, A// As s= AY /As is the average
rate of change of / with respect to the
distance As between C and C + AC.
Apart from infinitesimals of higher order,
An
— = COS \f/.
As Y'
where An denotes the distance from C to C + AC along the nor-
mal to C at (x, y), and \l/ is the angle between An and As; hence,
dn/ds = cos ^. Therefore,
(42-2) ' «»4f.*!»^co8*
' as an ds an
This relation shows that the derivative of / in any direction may
be found by multiplying the derivative along the normal by the
cosine of the angle ^ between the particular direction and the
normal. This derivative in the direction of the normal is called
the normal derivative of /. Its numerical value obviously is the
maximum value that df/ds can take for any direction. In applied
mathematics the vector in the direction of the normal, of magni-
tude df/dn, is called the gradient.
Example. Using (42-1), find the value of a that makes df/ds a
maximum, considering x and y to be fixed. Find the expression for
this maximum value of df/ds.
§42 PARTIAL DIFFERENTIATION
Since df/ds = /* cos a + /„ sin a,
d /df\
145
The condition for a maximum requires that
tanai =^, or a, = tan'1^-
Using this value of «i,
j: °r
f,
The relation (42-2) can be derived
directly by use of this expression for
df/dn. If a (Fig. 32) gives any di-
rection different from the direction
given by QJI, then
~ = /, cos a + fy sin a.
But a = ai — \l/1 so that
C+JC
FIG. 32.
.2. = /j.(cos «i cos ^ -f sin ai sin
cos \/> — cos «i sn
Since
cos ai =
^ COS ^ + /a
^fr!cos* /v/5+^
i«9 I J»2
= cos ^ = Vfi +fi cos \
= -T21 COS
dn
PROBLEMS
1. Find the directional derivative of f(x, y) = x*y + sin xy at
(l,7r/2), in the direction of the line making an angle of 45° with the
146 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §43
2. Find
dn
dx
if x — r cos 8, y = r sin 0, and / is a function of the variables r and 0.
3. Find the directional derivative of f(x, y) = a;3?/ + ev* in the
direction of the curve which, at the point (1, 1), makes an angle of
30° with the #-axis.
4. Find the normal derivative of
f(x, y) = x* + y*.
43. Tangent Plane and Normal
Line to a Surface. It will be re-
called that
Ax + By + Cz = D
. 33. . .
is the equation of a plane, where
the coefficients A, B, and C are called the direction components
of the normal to the plane. If a, (3, and y (Fig. 33) are the direc-
tion angles made by the normal to the plane from the origin, then
cos a =
Therefore,
A
** cos 8 =
B
VA2 + B2 + C2
cos y = — 7=
c
\/A* + B2 + C2
+ B2 + C2
cos a : cos ft : cos 7 = A:B:C.
If the plane passes through the point (XQ, yo, ZQ), its equation can
be written as
A(x - x0) + B(y - T/O) + C(z - ZQ) = 0.
There is also a normal form for the equation of a plane, entirely
analogous to the normal form for the equation of the straight
line in the plane. This form is
or
a; cos a + 2/ cos /3 + 2 cos 7 = p,
B
+ B* +
+ #2 + c2
+ B* + C2
D
+ B2 + C2'
§43
PARTIAL DIFFERENTIATION
147
in which p = D/^/A2 + J52 + C2 is the distance from the origin
to the plane.
Consider a surface denned by z = f(x, y), in which x and y
are considered as the independent variables. Then,
(43-1)
If #o and t/o arc chosen, z0 is determined by z = /(x, y
Ax = x — x0 and Ay = y — y0, and denote cfe by z — z0.
(43-1) becomes
. Let
Then
(43-2) z - z0 =
(.ro,
(X - X0) +
Jo, ?/o)
- 2/0),
which is the equation of a plane. If this plane is cut by the
plane x = x0, the equation of the line of intersection is
•o, yo)
and this is the tangent line to the curve z = /(xo, y) at the point
(x0, 2/0, z0). Similarly, the line of intersection of the plane defined
by (43-2) and the plane y = 2/0 is the tangent line to the curve
z = /(x, T/O) at (x0, 2/0, z0). The plane defined by (43-2) is called
the tangent plane^to the surface z = /(x, y) at (x0, 2/0, z0).
The direction cosines of the normal to this plane are propor-
tional to
dx
dy
-1.
The equation of the normal line to the plane (43-2) at (x0, 2/0, z0)
is therefore
(43-3)
x — x0
y ~
Z — Zp
dJL
dx
df_
, 2/0)
(so, 2/0)
This line is defined as the norrrial to the surface at (#o, 2/o, z0).
Figure 34 shows the difference between dz = RP' and Az = RQ.
P(%o, 2/o, z0) is the point of tangency and R(XO + Ax, t/0 + Ay, z0)
is in the plane z = z0. PP7 is the tangent plane.
148 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §43
In case the equation of the surface is given in the form
F(x, y, z) = 0,
the tangent plane and the normal line at (x0) y0, z0) have the
respective equations
t \ -L. dF ( \
(X -— Xo; -f- ^-7 (y yo)
(xo, in, zo)
(xo, yo, z0)
dz
(n,
(z -
and
(43-5)
x XQ y y$
dF\ = dF\
^•^|(xo, i/o, zo) ®y\ (%o, yo, 20)
z —
(xo,yo,«o)
These equations follow directly from (41-2).
FIG. 34
Example 1. At (6, 2, 3) on the surface z2 + ?/2 + ^ = 49, the
tangent plane has the equation
(x - 6) +
(6, 2, 3)
The normal line is
(6, 2, 3)
(y -
(6, 2, 3)
(z - 3) = 0
6x + 2y + 3z = 49.
s - 6 __ y - 2 __ z - 3
~~ ~~ ~~
Example 2. For (2, 1, 4) on the surface z — x2 + y2 — 1, the tangent
plane is
z - 4 = 2x
(2, 1)
(x - 2) +
or
2y - z = 6.
§44 PARTIAL DIFFERENTIATION 149
The normal line is
x - 2 _ y - 1 z - 4
~T~ - 2 - -r
PROBLEMS
1. Find the distance from the origin to the plane x + y + z = 1.
2. Find the equations of the tangent plane and the normal line to
(a) 2x* + 3?/ + 4z2 = 6 at (1, 1, K);
(*) T + ?-S ~ lat(4>3>8)'
(<0 ^2 + |i + ^ = 1 at (*o, 2/o, *o);
(d) z2 + 2*/2 -- z2 = 0 at (1, 2, 3).
3. Referring to (43-4), show that
dF OF dF
cosa:cos0-cos7 '^Wl?
where cos a, cos j8, cos 7 are direction cosines of the normal line.
4. Show that the sum of the intercepts on the coordinate axes of any
tangent plane to x^ + yW + z^ = a1/2 is constant.
44. Space Curves. It will be recalled that a plane curve C
whose equation is
(44-1) y = f(x)
can be represented in infinitely many ways by a pair of parametric
equations
(44-2) x = x(t),
y = y(t)
so chosen that when the independent variable t runs continuously
through some set of values ti < t < t% the corresponding values
of x and y, determined by (44-2), satisfy (44-1).
For example, the equation of the upper half of a unit circle
with the center at the origin of the cartesian system,
y =
can be represented parametrically as
x = cos tj
y — sint, (0 ^ t ^ TT),
150 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §44
or
or
x = 2t,
y = A/i .- 4*2, (0 <; i :£ M).
Similarly, a space curve C can be represented by means of a
set of equations
x = x(t\
(44-3) < T/ -
so selected that when t runs through some set of values the
coordinates of the point P(x, y, z), defined by (44-3), trace out
the desired curve C.
FIG. 35.
i
It will be assumed that the functions in (44-2) and (44-3)
possess continuous derivatives with respect to t, which implies
that the curve C has a continuously turning tangent as the point
P moves along the curve.
Let P(xQ, yo, ZQ) (Fig. 35) be a point of the curve C defined by
(44-3) that corresponds to some value U of the parameter ty
and let Q be the point (XQ +' A#, yo + Ay, z0 + Az) that cor-
responds to t = to + A£. The direction ratios of the line PQ
§45 PARTIAL DIFFERENTIATION 151
joining P and Q are
Ax A?/ Az __ Ax^A^Az
Ac Ac* Ac A£* A£* A2
If A£ is allowed to approach zero, Ax, At/, and Az all tend to zero,
so that the direction ratiohs of the tangent line at P(XQ, yQ, z0)
are proportional to (dx/df)^t0'(dy/dt)t^to-(dz/d()t^to- Hence, the
equation of the tangent line to C at P is
x — XQ _ y --• j/o _ z — zp
where primes denote derivatives with respect to £.
Example. The equation of the tangent line to the circular helix
x — a cos t,
y = a sin £,
z = a£,
at t - 7T/6, is
a
~ 2 ~2 a
The element of arc c?s is given by
(da)2 = (dz)2 + W + (^)2,
so that the length of a space curve C can be calculated from
rv2
. ,
The length of the part of the helix between the points (a, 0, 0) and
(0, a, 7ra/2) is
45. Directional Derivatives in Space. There is no essential
difficulty in extending th>e results of Sec. 42 to any number of
variables. Thus*, if u — f(x, y, z) is a suitably restricted func-
tion of the independent variables x, y, and z, then the directional
derivative along a space curve whose tangent line at some point
P(x, y, z) (Fig. 35) has the direction cosines cos (x, s), cos (y, s),
152 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §45
and cos (z, s) is
du du f N . du f N , du f ,
^ = - cos (x, s)+- cos (i,, .) + - cos (z, ,)
The magnitude of the normal derivative to the surface u = const.
is given by
'
The vector that is normal to the surface u — const, and whose
magnitude is du/dn is called the gradient of u.
PROBLEM
1. Find the equation of the tangent line to the helix
x — a cos t, y — a sin t, z = a£,
at the point where t = Tr/4. Find the length of the helix between the
points t = 0 and t — 7T/4.
2. Find the directional derivative of / = xyz at (1, 2, 3) in the
direction of the line that makes equal angles with the coordinate axes.
3. Find the normal derivative of / = x2 + y2 + z2 at (1, 2, 3).
4. Show that the square root of the sum of the squares of the
directional derivatives in three perpendicular directions is equal to
the normal derivative.
6. Express the normal derivative (45-1) in spherical and cylindrical
coordinates, for which the equations of transformation are
(a) x = r sin 0 cos <p, y — r sin 6 sin <p, z = r cos 6;
(b) x = r sin 0, y = r cos 0, z = z.
6. What is the direction of the curve x = t, y — t2, z — £3 at the
point (1, 1, 1)?
7. Show that the condition that the surfaces f(x, y, z) =0 and
g(x, y,z) =0 intersect orthogonally is that
, . , = 0
dx dx dy dy dz dz
8. Show that the surfaces
xyz = 1
intersect at right angles.
x2 y2 z2
xyz = 1 and -^ + -% - f =
§46 PARTIAL DIFFERENTIATION 153
9. Find the angle between the normals to the tangent planes to
the surfaces x* + y2 + zz = 6 and 2z2 + 3?/2 + z* = 9 at the point
(1, 1, 2).
10. Show that the direction of the tangent line to the curve of inter-
section of the surfaces f(x, y, z) = 0 and g(xt y, z) = 0 is given by
a
cos a:cos j3:cos 7 =
/•
A />
Hint: Let (XQ, T/O, ^o) be a point on the curve of intersection, and find
the line of intersection of the tangent planes to the surfaces at the
point (XQ, 7/0, Zo).
46. Higher Partial Derivatives. The partial derivatives f*i9
/v * * * y f* °^ f(Xl> xz, ' ' ' j #n) are functions of x\, x%, • • - ,
xn and may have partial derivatives with respect to some or all
of these variables. These derivatives are called second partial
derivatives of f(xiy xz, ' ' ' , xn). If there are only two inde-
pendent variables x and t/, then f(x, y) may have the second
partial derivatives
df\ _ d2/ _
Jx) ^ dx* ^ ***'
^ JXV)
dy \dx/ dy dx
dx \dy/ dx dy
d_(V\s^sf
dy \dy/ dy2 v'
It should be noticed that fxy means that df/dx is jBrst found and
then — ( — ) is determined, so that the subscripts indicate the
dy \dx/
order in which the derivatives are taken. In
a2/ _ a (ef\
dy dx dy \dx/
the order is in keeping with the meaning of the symbol, so that
the order appears as the reverse of the order in which the deriva-
tives are taken.
It can be proved* that, if f^ and/^ are continuous functions
of x and yf then /^ = fyx, so that the order of differentiation is
* See SOKOLNIKOFF, I. S., Advanced Calculus, Sec. 31.
154 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §46
immaterial. Similarly, when third partial derivatives are
found, fxyx = fxxy = fyxx and f^y = fvxy = fvyx, if these deriva-
tives are continuous.
Example. If f(x, y) = e**, then
/* = ye*", fv = £6**, /«,« yV*,
/«* = /** = ^xv(^ + 1), /yy = tfe*"-
PROBLEMS
1. Verify that ^- = ^- for
J 60; d?/ cty d#
(a) / = cos xy2, (6) / = sin2 x cos y, (c) f = ey/x.
2. Prove that if
v d*f d2f
(a) /(*, y) = log (»» + t/2) + tan-1 J then -^ + ^ = 0;
(W /<*, y, «) - W + 2/2 + ^)~^, then g + g + g = 0.
3. If u = x* + y* and \X = * + 3j' find ~ and ~-
J (y = 2s — t, ds* dt2
(x = r cos 0, d2u 32u
4. If u = /(x, y) arid j gin ^ find ^-2 and ^-
5. Use the results obtained in Prob. 6(6), Sec. 41, in order to show
that ~ - u(W - u2)/(u* + v2)*. Find ^- and |~-
2 ^ // V • / l
6. If -a? = ^(x, ?/), where x — x(u, v), y = t/(w, ?;), dx/du — dy/dv,
and dx/dv = —dy/dUj show that
_
dw2 " ^2 ~ \dx2
7. Show that the expressions
dz* . /dz\* , a2^ d2z
upon change of variable by means of x = r cos 6 and i/ = r sin ^ become
T/ /^Y , 1 /^A2 A T/ d20 , 1 ^22 1 ^
Fl = \dr) + ^ (do) and F2 - 5? + F2^2
8. If F = /(a; + ct) + g(x — ct}} where / and ^ are any functions
possessing continuous second derivatives, show that
_ ,
" c
§47 PARTIAL DIFFERENTIATION 155
9. Show that if x — er cos 0 and y = er sin 8, then
?!Z . ^!Z _ -2 /d2]/ , d!T\
dx* + a?/2 ~'e A^2 a* V
10. If 7i(x, 2/, 2) and y2(z, T/, 2) satisfy the equation
show that
satisfies the equation
d2 d2 d2
47. Taylor's Series for Functions of Two Variables. This sec-
tion contains a formal development of a function of two variables,
/(x, t/), in a series analogous to the Taylor's series development
of a function of a single variable. It is assumed that the series
obtained here converges to the value of the function f(xt ?/),
but the analysis of the conditions under which this convergence
will occur is too involved to be discussed in this book.
Consider /(x, y), which is a function of the two variables x and
T/, and let it have continuous partial derivatives of all orders.
Let
(47-1) x = a + at and y = b + pt,
where a, 6, a, and p are constants and t is a variable. Then
(47-2) /(x, y) = /(a + at,'b + pt) = F(t).
If F(t) is expanded in Maclaurin's series, there results
F"(Q) F"'(Q)
. L W *2 _L L W *3 _1_ . . .
(47-3)
From (47-2) and (47-1), it follows that
= /.(»,»)« +fy(x,y)P-
Then,
156 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §47
and
dx
+ fyyx(x, y}p] -
3fxyy(x, y)a
Higher order derivatives of F(t) can be obtained by continuing
this process, but the form is evident from those already obtained.
Symbolically expressed,
°toa? + ^d»»'
Then,
n_A dnf _ n o^/
where
n
-r ~r!(n_r)!
Since £ = 0 gives x = a and ?/ = 6, it follows that
F(0) = /(a, 6), F'(0) = afx(a, 6) + )8/y(a, 6), - • • .
Substituting these expressions in (47-3) gives
*XO = /(», y) = /(«, fc) + W»(a, &) + fify(a, &)]«
Since at = a; — a and # = y — 6, the expansion becomes
(47-4) /(x, 2/) = /(a, 6) +/«(a, 6)(* - a) +/,(a, 6)(y - 6)
+ A [/*»(<*, b)(» - «)2 + 2/^(a, 6)(x - a)(y - 6)
§47 PARTIAL DIFFERENTIATION 157
This is Taylor's expansion for a function f(x, y) about the point
(a, b).
Another form that is frequently used is obtained by replacing
(x — a) by h and (y — b) by k, so that x = a + h and y = b + fc.
Then,
(47-5) f(a + h,b + k) = /(a, 6) + /,(a, V)h + /v(a, 6)fc
+ £j r/~(a, &)^2 + 2/*,(a, b)hk + /w(a, 6)fc2] + • • • .
This formula is frequently written symbolically as
/(a + h, b + fc) = /(a, 6) + (h ^ + *
Example. Obtain the expansion of tan"1 - about (1,1) up to the
y
third-degree terms. Here, f(x, y) = tan"1 -> so that
x
f(x, y) = tan-1 -> /(I, 1) = tan"1 1=7;
/»(«i y) =
xxfe y) =
/vv(«> y) = (^2 _|_ yaj
Then,
PROBLEMS
1. Obtain the expansion for x^/2 + cos xy about (1, Tr/2) up to the
third-degree terms.
158 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §48
2. Expand f(x, y) = exv at (1, 1), obtaining three terms.
3. Expand ex cos y at (0, 0) up to the fourth-degree terms.
4. Show that, for small values of x and y,
and
e* sin y = y + xy (approx.),
y2
'• log (1 + y) = y + xy - ^ (approx.).
5. Expand f(x, y) = x*y + x2y + 1 about (0, 1).
6. Expand (1 — x2 — y2)^ about (0, 0) up to the third-degree
terms.
7. Show that the development obtained in Prob. 6 is identical with
the binomial expansion of [1 — (x2 -f y2)]}/^.
48. Maxima and Minima of Functions of One Variable. A
function f(x) is said to have a maximum at x = a, if
and
A+ s /(a + h) - /(a) < 0,
A- = /(a - A) - /(a) < 0,
for all sufficiently small positive values of h. If A+ and A"~ are
both positive for all small positive values of h, then f(x) is said
to have a minimum at # = a.
It is shown in the elementary calculus that, if the function
f(x) has a derivative at x = a, then the necessary condition for a
maximum or a minimum is the
vanishing of f'(x) at the point
x = a. Of course, the function
f(x) may attain a maximum
or a minimum at x = a with-
out having /'(a) = 0, but this
^x can occur only if /'(#) ceases to
exist at the critical point (see
Fig. 36).
Let it be supposed that f(x) has a continuous derivative of
order n in some interval about the point x = a. Then it follows
from Taylor's formula that
a
FIG. 36.
§48 PARTIAL DIFFERENTIATION , 159
where 0 < 61 < 1, and
A- B /(a - h) - /(a)
where 0 < 02 < 1- I^et it be assumed further that the first
n — 1 derivatives of f(x) vanish at x = a but that /(n) (a) is not
zero. Then
n\
and
Since f(n)(x) is assumed to be continuous in some interval
about the point x = a, f(n)(a + Bik) and f(n}(a - eji) will
have the same sign for sufficiently small values of h. Conse-
quently, the signs of A+ and A~ will be opposite unless n is an
even number. But if f(x) is to have a maximum or a minimum
at x — a, then A+ and A~ must be of the same sign. Accordingly,
the necessary condition for a maximum or a minimum of f(x)
at x = a is that the first non-vanishing derivative of f(x), at
x — a, be of even order. Moreover, since both A4" and A~
are negative if f(x) is a maximum, it follows that/(n)(a) must be
negative. A similar argument shows that, if /(#) has a minimum
at x = a, then the first non-vanishing derivative of f(x) at x = a
must be of even order and positive.
If the first non-vanishing derivative of f(x) at x = a is of odd
order and/" (a) = 0, then the point x = a is called a point of inflection.
Example. Investigate f(x) — x5 — 5#4 for maxima and minima.
Now,
f(x) - 5z4 - 20z3,
which is zero when x — 0 and x = 4. Then,
f"(x) = 20x3 - 60s2, /"(O) = 0, /"(4) * 320;
/'"(a?) = 60s2 - 120s, /'"(O) = 0;
f™(x) = 120s - 120, /IV(0) = -120.
Since /"(4) > 0, /(4) = -256 is a minimum; and since /IV(0) < 0,
/(O) = 0 is a maximum.
160 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §49
PROBLEMS
1. Examine the following for maxima and minima:
(a) y = x* - 4x* + 1 ;
(6) y = x*(x - 5)2;
(c) y — x + cos x.
2. Find the minimum of the function y — xx, where x > 0.
Hint: Consider the minimum of log y.
3. Show that x = 0 gives the minimum value of the function
y — ex + 6~z + 2 COS £.
4. Find maxima, minima, and points of inflection, and sketch the
curves, for the following:
(a) y — 3x -f 4 sin x + sin 2x;
(b) y — 3x — 4 sin x + sin 2z;
(c) i/ = 6z + 8 sin x + sin 2x.
6. Find the maximum and minimum values of the function
y — x sin x + 2 cos x.
6. Find maxima, minima, and points of inflection, and sketch the
curves, for the following:
(a) y = zlogz;
(b) x5 - (y - x2)* = 0.
49. Maxima and Minima of Functions of Several Variables.
A function of two variables f(x, y) is said to have a maximum at
(a, £>), if /(a + h, b + k) - /(a, 6) < 0 for sufficiently small
positive and negative values of h and ft, and a minimum, if
/(a + h, b + k) - /(a, 6) > 0.
Geometrically, this means that when the point (a, 6, c) on the
surface z — f(x, y) is higher than all neighboring points, then
(a, 6, c) is a maximum; and when (a, b, c) is lower than all
neighboring points, it is a minimum point. At a maximum or a
minimum point (a, 6, c) the curves in which the planes x = a and
y = b cut the surface have maxima or minima. Therefore,
f*(a, 6) = 0 and /y(a, 6) = 0. The conditions fx = 0 and /y = 0
can be solved simultaneously to give the critical values.
The testing of the critical values for maxima and minima is
more difficult than in the case of functions of one variable.
However in many applied problems the physical interpretation
§49 PARTIAL DIFFERENTIATION 161
will determine whether or not the critical values yield maxima
or minima or neither. An analytical criterion can be established
for the case of two variables in a manner analogous to the method
used for one variable. By the use of Taylor's expansion, it can
be shown that if /x(a, b) = 0 and fv(a, 6) = 0, then /(a, 6) is a
maximum if
D ^ &(a, b) - /«,(a, b}fyy(ay b) < 0
with
/**(«, 6) < 0 and fyy(a, 6) < 0,
and a minimum if
D**fi,,(a,b) -/«,(a,6)/w(a,6) <0
with
fxx(a, b) > 0 and fyy(a, b) > 0.
In case
fly(a, b) - fxx(a, 6)/w(a, 6) > 0,
/(a, fc) is neither a maximum nor a minimum. If
&(M) ~ /~(«, &)/**(«, &) =0,
the test gives no information, just as/"(z) = 0 gives no criterion
in the case of one variable.
These considerations can be extended to functions of more than
two variables. Thus, in the case of a function f(x, y, z) of three
variables,
f , o, £ = 0, % = 0
dx ' dy ' dz
is the necessary condition for a maximum or a minimum.
Example 1. A long piece of tin 12 in. r
wide is made into a trough by bending up \
the sides to form equal angles with the
base (Fig. 37). Find the amount to be ^t-\.
bent up and the angle of inclination of the /2-2x
sides that will make the carrying capacity 0,_
Jb IG. o7.
a maximum.
The volume will be a maximum if the area of the trapezoidal cross
section is a maximum. The area is
A = I2x sin 6 - 2x2 sin 0 + x2 sin 9 cos 0;
for 12 — 2x is the lower base, 12 — 2x + 2x cos 6 is the upper base,
162 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §49
and x sin 6 is the altitude. Then,
33- = I2x cos 6 - 2z2 cos 6 + x2 cos2 6 - x2 sin2 0
O0
= z(12 cos 6 - 2z cos 0 + z cos2 0 - z sin2 0)
and
|~ = 2 sin 0(6 - 2z -f x cos 0).
dA/dx = 0 and 3A/dB — 0, if sin 0 = 0 and z = 0, which, from physical
considerations, cannot give a maximum.
There remain to be satisfied
6 - 2x + x cos 0 = 0
and
12 cos 0 - 2x cos 6 -f x cos2 6 - x sin2 0 = 0.
Solving the first equation for x and substituting in the second yield,
upon simplification,
cos 0 = J-6 or 0 = 60°, and x = 4.
Since physical considerations show that a maximum exists, x = 4 and
0 = 60° must give the maximum.
Example 2. Find the maxima and minima of the surface
Now,
d% x dz
which vanish when x = y = 0. But
dx2 a2c dy2 b2c dx dy
Hence, D = l/a262c2and, consequently, there is no maximum or minimum
at x = y = 0. The surface under consideration is a saddle-shaped
surface called a hyperbolic paraboloid. The points for which the first
partial derivatives vanish and D > 0 are called minimax. The reason
for this odd name appears from a consideration of the shape of the
hyperbolic paraboloid near the origin of the coordinate system. The
reader will benefit from sketching it in the vicinity of (0, 0, 0).
PROBLEMS
1. Divide a into three parts such that their product is a maximum.
Test by using the second derivative criterion.
§60 PARTIAL DIFFERENTIATION 163
2. Find the volume of the largest rectangular parallelepiped that can
be inscribed in the ellipsoid
4. -4- - - i
a2"1" 62 "^c2
3. Find the dimensions of the largest rectangular parallelepiped that
has three faces in the coordinate planes and one vertex in the plane
4. A pentagonal frame is composed of a rectangle surmounted by an
isosceles triangle. What are the dimensions for maximum area of the
pentagon if the perimeter is given as P?
5. A floating anchorage is designed with a body in the form of a right-
circular cylinder with equal ends that are right-circular cones. If the
volume is given, find the dimensions giving the minimum surface area.
6. Given n points P% whose coordinates are (#», 7/t, zt), (i — 1,2,
• • • , n). Show that the coordinates of the point P(x, y, z), such that
the sum of the squares of the distances from P to the P% is a minimum,
are given by
(1 n 1 n^ i n» \
nS^S^S*)'
50. Constrained Maxima and Minima. In a large number of
practical and theoretical investigations, it is required that a
maximum or minimum value of a function be found when the
variables are connected by some relation. Thus, it may be
required to find a maximum of u = f(x, yt z), where x, y, and z
are connected by the relation <p(x, y, z) = 0. The resulting
maximum is called a constrained maximum.
The method of obtaining maxima and minima described in
the preceding section can be used to solve a problem of con-
strained maxima and minima, as follows: If the constraining
relation <p(x, y, z} = 0 can be solved for one of the variables,
say z, in terms of the remaining two variables, and if the resulting
expression is substituted for z in u = f(x, y, 2), there will be
obtained a function u = F(xt y). The values of x and y that
yield maxima and minima of u can be found by the methods of
Sec. 49. However, the solution of <p(x, y, z) = 0 for any one
of the variables may be extremely difficult, and it is desirable to
consider an ingenious device used by Lagrange.
164 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §50
To avoid circumlocution the maximum and minimum values
of a function of any number of variables will be called its extremal
values. It follows from Sec. 49 that the necessary condition
for the existence of an extremum of a differentiable function
f(xi, #2, • • • , £n) is the vanishing of the first partial derivatives
of the function with respect to the independent variables Xi,
Xz, ' ' * , %n- Inasmuch as the differential of a function is
defined as
it is clear that df vanishes for those values of x\, Xz, • ' ' , xn
for which the function has extremal values. Conversely, since
the variables xt are assumed to be independent, the vanishing
of the differential is the necessary condition for an extremum.
It is not difficult to see that, even when some of the variables
are not independent, the vanishing of the total differential is
the necessary condition for an extremum. Thus, consider a
function
(50-1) u = f(x, y, z),
where one of the variables, say z, is connected with x and y by
some constraining relation
(50-2) <p(x, y, z) = 0.
Regarding x and y as the independent variables, the necessary
conditions for an extremum give du/dx = 0 and du/dy = 0, or
du ^ df =
dx dx "*" dz dx '
dy dy ~~ dz dy
Then the total differential
du , , du j df , . df . . dfdz , , dz
- ~ -
and since the expression in the parenthesis is precisely dz, it
follows that
(5°-3) ^ + ^ + ^0-
§60 - PARTIAL DIFFERENTIATION 165
The total differential of the constraining relation (50-2) is
5*
£*-«•
Let this equation be multiplied by some undetermined mul-
tiplier X and then added to (50-3) . The result is
Now, if X is so chosen that
(50-5)
f\ i^ **• »\
dx dx
, y,
= o,
then the necessary condition for an extremum of (50-1) will
surely be satisfied.
Thus, in order to determine the extremal values of (50-1),
all that is necessary is to obtain the solution of the system of
Eqs. (50-5) for the four unknowns x, y, z, and X. The multiplier
X is called a Laqrangian multiplier.
Example 1. Find the maximum and the minimum distances from
the origin to the curve
5z2 + fay + 5?/ -8 = 0.
The problem here is to determine the extremal values of
f(x, y) = xz + y2
subject to the condition
<p(x, y) SE 5z2 + fay + 5i/2 - 8 = 0.
Equations (50-5) in this case become
2x + \(Wx + 6y) = 0,
2y + X(6z + 102/) = 0,
5z2 + fay + 5y2 - 8 = 0.
Multiplying the first of these equations by y and the second by x and
166 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §50
then subtracting give
6X(?/ - a:2) = 0,
so that y = ±x. Substituting these values of y in the third equation
gives two equations for the determination of x, namely,
2z2 = 1 and x2 = 2.
The first of these gives / = x1 + y2 = 1, and the second gives/ = x2
+ 2/2 = 4. Obviously, the first value is a minimum, whereas the
second is a maximum. The curve is an ellipse of semiaxes 2 and 1
whose major axis makes an angle of 45° with the z-axis.
Example 2. Find the dimensions of the rectangular box, without a
top, of maximum capacity whose surface is 108 sq. in.
The function to be maximized is
f(xt y, z) 35 xyz,
subject to the condition
(50-6) xy + 2xz + 2yz = 108.
The first three of Eqs. (50-5) become
( yz + \(y + 2z) = 0,
(50-7) 4 xz + \(x + 2z) = 0,
(xy + \(2x + 2y) =0.
In order to solve these equations, multiply the first by x, the second
by ?/, and the last by z, and add. There results
\(2xy + 4xz + 4?yz) + 3xyz = 0,
or
2xz + 2yz) + %xyz = 0.
Substituting from (50-6) gives
108X + %xy* = 0,
or
\ _ _ *y*
A ~ 72'
Substituting this value of X in (50-7) and dividing out common factors
give
1 - ^ (y + 2z) = 0,
i - ^ (* + ao - o,
1 - (2* + 2y) - 0.
§61 PARTIAL DIFFERENTIATION 167
From the first two of these equations, it is evident that x = y. The
substitution of a: = y in the third equation gives z = 18/y. Substitut-
ing for y and z in the first equation yields x = 6. Thus, x = 6, y = 6,
and 3 = 3 give the desired dimensions.
PROBLEMS
1. Work Probs. 1, 2, and 3, Sec 49, by using Lagrangian multipliers.
2. Prove that the point of intersection of the medians of a triangle
possesses the property that the sum of the squares of its distances from
the vertices is a minimum.
3. Find the maximum and the minimum of the sum of the angles
made by a line from the origin with (a) the coordinate axes of a cartesian
system; (b) the coordinate planes.
4. Find the maximum distance from the origin to the folium of
Descartes x3 + y3 — 3<m/ = 0.
5. Find the shortest distance from the origin to the plane
ax + by + cz = d.
51. Differentiation under the Integral Sign. Integrals whose
integrands contain a parameter have already occurred in the
first chapter. Thus, the length of arc of an ellipse is expressi-
ble as a definite integral containing the eccentricity of the ellipse
as a parameter. *
Consider a definite integral
(51-1) *(«) = £'/(*, «) **,
in which the integrand contains a parameter a and where UQ and
ui are constants. As a specific illustration, let
7T
/*2
<p(a) = I sin ax dx.
Jo
In this case the indefinite integral
f . cos ax ~
F(x, ot) — I sin ax dx = h C
is a function of both x and a ; but, upon substitution of the limits,
there appears a function of a. alone, namely
IT T
, . P2 . , cos ax\2 I/- TrcA
via) = I sm ax ax = = - I 1 — cos -^ I-
Jo . a |o a \ 2 /
* See Sec. 14.
168 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §61
Frequently, it becomes necessary to calculate the derivative
of the function <p(a) when the indefinite integral is complicated
or even cannot be written down explicitly. Inasmuch as the
parameter a is independent of x, it appears plausible that in
some cases it may be permissible to perform the differentiation
under the integral sign, so that one can use the formula
dtp _ J Ul df(Xj a) ,
da Juo da
This formula turns out to be correct if /(x, a) and d/(z, a) /da
are continuous functions in both x and a. Thus, forming the
difference quotient with the aid of (51-1),
Juo
- f(x, «)
v ' Aa Jwo Aa
Now the limit, as Aa —> 0, of the left-hand member of (51-2)
is precisely dy/da, whereas the limit of the expression under the
integral sign is df/da. Hence, if it is permissible to interchange
the order of integration and calculation of the limit, one has
-
da
The restrictions imposed on the function f(x, a) can be shown
to be sufficient to justify the inversion of the order of these
operations.
Suppose next that the limits of integration u\ and UQ are func-
tions of the parameter a, so that
In this case, one can proceed as follows : Let
ff(x,a)dx = F(x,a)
so that
(51-4) ~ = f(x, a).
Then,
(51-5) <f>(a) = f"1™ f(x, a) dx = F(x, a)
Juo(a)
i, a) - F(tt0, a).
§61 PARTIAL DIFFERENTIATION 169
Assuming the continuity of all the derivatives involved, one
can write*
, g)
_ _ __ __
da du\ da da du$ da da
which, upon making use of (51-4) and (51-5), becomes
~ , a)]
- A"., «) p + £ fU>M f(x, a) dx.
a da da Jwo(«)
The partial derivative appearing in this expression means that
the differentiation is to be performed with respect to a, treating
UQ and u\ as constants. Hence, making use of (51-3),
/r, ^ d<p ., . du\ „, . duo . Cui(a} df(xy a) ,
(51-6) = f(Ul, a)-- f(u0, a) + J--l dx.
This formula is known as the formula of Leibnitz, and it
specializes to (51-3) when HI and UQ arc independent of a. The
validity of this formula can be established under somewhat less
restrictive hypotheses,! but the limitations imposed on the func-
tion f(x, a) in the foregoing discussion are usually met in prob-
lems arising in applied mathematics.
d- r2a -—
Example 1. Find -T-, if #(a) = J_a2 e "2 dx.
Then
/2a 9rl> _ r!
-«» ^3* e ""' (JX
Example 2. Formula (51-3) is frequently used for evaluating definite
integrals. Thus, if
(a) = J0 log (1 + a cos x) dx,
* See Sec. 39.
t See SOKOLNIKOFF, I. S., Advanced Calculus, Sec. 39, p. 121.
170 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §51
then
=lxfo(l- 1+acosJ
cos
1 + a cos #
1 r , 1 /.,«-! . ,<* + 1\1
= - TT H -- 7 — : -- ( sin"1 -- sm"1 - )
al Vl -«2\ 1 -a 1 + a/J
=
" a
Therefore,
(D\CX.) ==7T I I — — 7 — :==:-
J \a a\/l - az
i + vr^s
g a + log
\
or
<p(a) = TT log (I + Vl ~ «2) + c.
But, when a = 0,
Hence,
0 = TT log 2 + c and c = — TT log 2,
and ___ _
, . , /i + Vi - «2\
^(a) - TT log ^—T-^ - J.
PROBLEMS
J/2
0 sin ao; d^ by using the Leibnitz formula,
and check your result by direct calculation.
2. Find dp/da, if <p(a) = J Q (1 — a cos a:)2 dx.
r«a a;
3. Find d<p/da, if ^(a) = J0 tan-1 ^ ete.
4. Find dp/dot, if <p(» = JQ tan (x "" a) ^a;-
5. Find d(p/dxy if ^(o?) = J Q V^ ^-
/*7T fa
6. Differentiate under the sign and thus evaluate JQ -7 — — - ^
. C* dx 7T .. 0 . .
by using L - = —5 - r> if a2 > 1.
J b Jo a — cos # a2 — 1
7. Show that
fT log (1 - 2a cos 3 + a2) dx = 0, if a2 ^ 1
= ir log a2, if aa ^ 1,
§61 PARTIAL DIFFERENTIATION 171
8. Verify that
is a solution of the differential equation
jg3 + k*y = /(&)-
where A; is a constant.
CHAPTER V
MULTIPLE INTEGRALS
It is assumed that the reader is somewhat familiar with the
problem of calculating the volumes of solids with the aid of double
and triple integrals and has some facility in setting up such
integrals. The first three sections of this chapter contain a
brief summary of some basic facts concerning double and triple
integrals, preparatory to the development of the expressions for
the volume elements in spherical and cylindrical coordinates.
These expressions are used frequently in applied mathematics
and are seldom included in the first course in calculus. A brief
discussion of surface integrals is also given here.
First, it may be well to recall the definition of the simple
integral J*£ f(x) dx. Let the function f(x) be continuous and
single valued for a < x < b. The interval (a, b) of the #-axis
is divided into n parts by the points a z= XQ, xi, x2, • • • ,
xn = b. Let A#» = Xi — #t_i, and let £t be a value of x such that
n
rct-i < £t ^ x^ Form the sum 23 /(£») A#t, and take the limit
t==i
of this sum as n — » <*> and all the Axl — > 0. Under the given
assumptions on f(x), this limit will exist, and it is defined as the
definite integral of f(x) over the interval (a, b) of the #-axis.
Thus,
lim X/({.) A*. ^ P /(*)**-
Geometrically, this integral can be interpreted as the area between
the curve y = f(x) and the x-axis included between the lines
x = a and x = b. The evaluation of the integral can often be
accomplished by the use of the following theorem.
FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS. // f(x)
is continuous in the interval a < x ^ b and G(x) is a function such
that dG/dx = f(x) for all values of x in this interval, then
fb f(x) dx = O(b) - G(a).
Ja
172
§52
MULTIPLE INTEGRALS
173
52. Definition and Evaluation of the Double Integral. The
double integral is defined and geometrically interpreted in a
manner entirely analogous to that sketched above for the simple
integral. Let f(x, y) be a continuous and single-valued function
within a region R (Fig. 38), bounded by a closed curve C, and
upon the boundary C. Let
the region R be subdivided
in any manner into n sub-
regions AjRi, A#2, ' ' ' , A.Rn
of areas A A i, A.A2, * * • ,AAn.
Let (£t, 7?t) be any point in
the subregion AjRt, and form
the sum
FIG. 38.
The limit of this sum, as n — > °o and all AAt — > 0, is defined as
the double integral of f(x, y) over the region R. Thus,
(52-1)
Km
f(x, y) dA.
The region R is called the region of integration, corresponding
to the interval of integration (a, b) in the case of the simple
integral. The integral (52-1) is sometimes written as
In order to evaluate the double integral, it will be simpler to
consider first the case in which the region R (Fig. 39) is a rec-
tangle bounded by the lines x — a, x = 6, y = c, y = d. The
extension to other types of regions will be indicated later.
Subdivide R into mn rectangles by drawing the lines x = Xi,
x = z2, • • • , x = Zn-i, y = yi, y = 2/2, • • • , # = y«-i.
Define Axt = zt — xt__i, where XQ = a and xn = 6, and define
Ay, = t/? — 2/j-i, where 2/0 = c and 7/w = d. Let A.Rt/ be the rec-
tangle bounded by the lines x = zt_i, # = xt, ?/ = y/-i, 2/ — 2/?-
Then, if the area of ARij is denoted by AA»y,
= Axt Ay/.
174 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §52
t = n, j *» m
Let (£t;, 7jt,) be any point of AB#. The sum S /(£*,-, ??»,) AAt,-
can be written as
t = n, /«"Wt
(52-2) j) /(**" ^»^ AXt A2/J''
This summation sign signifies that the terms can be summed for
i and j in any manner whatsoever. Suppose that the terms of
(52-2) are arranged so that all the rectangles A.Rti are used first,
y-
y=d
Jm-l
Vi
->JC
FIG. 39.
then all the rectangles A/^t2, then all the rectangles A72t3, etc.
This is equivalent to taking the sum of the terms for each row of
rectangles and then adding these sums. Then (52-2) can be
written
(52-3)
But
lim
(*.„' th,) Ax.].
>»5
*1) Ax» = Ja f(x> i
dx>
so that
where lim €, = 0. Moreover, | /(x, tij) dx is a function of 77,,
n— * « •/<*
§62 MULTIPLE INTEGRALS 175
say ^>0?,). Thus (52-3) becomes
;«1
= fcd v(y} dy + e(d - c) + e'
fd fb
~ I I f($> y} dx dy + c(d — c) + e ,
Jc Ja
in which lim e = 0 and lim ef — 0. Taking the limit as n — > oo
n— * « m— >• oo
and m — > oo gives
(52-4) Jfl f(x, y) dA = J" JT* /(x, y) do; dp.
The double integral is, therefore, evaluated by considering
f(Xj y) as a function of x alone, but containing y as a parameter,
and integrating it between x — a and x = 6 and then integrating
the resulting function of y between y — c and y = d. The right
member of (52-4) is known as an iterated integral, and (52-4)
establishes the relation between the double integral over the
rectangle R and an iterated integral over the same rectangle.
Similarly, by taking the sum of the terms in each column and
then adding these sums,
(52-5) f f(x, y) dA = f f* f(x, y) dy dx.
J K 4/O t/ C
In case (52-5) is used, f(x, y) is first considered as a function of y
alone and integrated between y = c
and y = d, and then the resulting
function of x is integrated between
x = a and x = b. Either (52-4) or
(52-5) can be used, but one of them
is frequently simpler in the case of a
particular f unction /(x, y).
Suppose R is, not a rectangle,
but a region bounded by a closed
curve C (Fig. 40) that is cut by any ' FIQ 4Q
line parallel to one of the axes in, at
most, two points. Let Bi and JS2 be the points of C having
the minimum and maximum ordinates, and let AI and A 2 be
the points of C having the minimum and maximum abscissas.
Let x = <?i(y) be the equation of B\A\B^ and x = <p*(y) be the
176 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §62
equation of BiA2B2. Then, in taking the sum of the terms by
rows and adding these sums, the limits for the first integration
will be <pi(y) and <p2(y), instead of the constants a and b. The
limits for the second integration will be 0i and /32, in which /3i is
the ^/-coordinate of BI and /32 is the y-coordinatc of B2. Then
(52-4) is replaced by
(52-6)
dA =
**
Similarly, if y = fi(x) is the equation of A\BiA^ y = fz(x) is
the equation of A\B^A^ a\ is the abscissa of Ai, and a2 is the
abscissa of A2, (52-5) is replaced by
(52-7)
/(*, y) dA =
/(,, y) dy dx.
y*b
Tn case R is a region bounded by a closed curve C that is cut
in more than two points by some
parallel to one of the axes, the
previous results can be applied to
subregions of R whose boundaries
satisfy the previous conditions.
By adding algebraically the inte-
grals over these subregions, the
double integral over R is obtained.
Example 1. Compute the value of
Fio 41
/i = f R y dA where R is the region in the first quadrant bounded by
the ellipse
?! 4_ yl - i
a2 "^ 62
Upon using (52-6) and summing first by rows,
n. . _ _
(Fig. 41).
dy
3
Using (52-7), one has
6 ,
/*ii /*~" v aj-
i, = r faV
Jo Jo
62
*3 /»0 /V2 " v a2-x2\
yd2/da; = Jo(2o )da:
§53
MULTIPLE INTEGRALS
111
It may be remarked that the value of /i is equal to yA, in which y is
the ^-coordinate of the center of gravity of this quadrant of the ellipse
and A is its area. Since A = irab/4,
/i ab^/3 46
$ = 1 = mb/l ** &T
Similarly, by evaluating 72 = f R x dA = a26/3,
3-7T
x = =
A 7ra&/4
which is the ^-coordinate of the center of gravity.
Example 2. Moment of Inertia. It will be recalled that the moment
of inertia of a particle about an axis is the product of its mass by the
square of its distance from the axis. If it is desired to find the moment
of inertia of a plane region about an axis perpendicular to the plane
of the region, the method of Sec. 52 can
be applied, where f(x, y) is the square of
the distance from the point (xy y) of the ^
region to the axis. Then
M =
dA\
^^
(l>°>
For example, let it be required to find
the moment of inertia of the area in the _
first quadrant (Fig. 42), bounded by the
parabola y2 — 1 — x and the coordinate
axes, about an axis perpendicular to the
xy-p\&ne at (1,0). The distance from any point P(xt y) to (1,0)
is r = \/(x - I)2 + y2- Therefore,
FIG. 42.
Evaluating this integral by means of (52-6) gives
M --
dy
53. Geometric Interpretation of the Double Integral. If
f(Xj y) is a continuous and single-valued function defined over the
region R (Fig. 43) of the xy-pl&ne, then z = f(x, y) is the equation
of a surface. Let C be the closed curve that is the boundary of R.
Using R as a base, construct a cylinder having its elements parallel
178 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §63
to the z-axis. This cylinder intersects z = f(x, y) in a curve T,
whose projection on the ;n/-plane is C. Denote by S the portion
of z = f(x, y) that is enclosed by T. Let R be subdivided as in
Sec. 52 by the lines x = xl, (i = 1, 2, • • • , n — 1), and y = y]y
(j = 1, 2, - - - , m - 1).
Through each line x = x%
pass a plane parallel to the
2/z-plane; and through each
line y = y/ pass a plane
parallel to the zz-plane.
The rectangle A72l?-, whose
area is AAtJ = AxtAyJ} will
be the base of a rectangular
^y prism of height /(£t?, r;t3),
whose volume is approxi-
mately equal to the volume
enclosed between the surface
and the #?/-plane by the
planes x = xt~i, x = ict,
2/ = 2/j-i, and i/ = j//. Then
the sum
FIG. 43.
(53-1)
"T"
<-T7-i
c* At/,-
gives an approximate value for the volume V of the portion of the
cylinder enclosed between z = f(xf y) and the xy-pl&ne. As
oo and m — > °o; the sum (53-1) approaches F, so that
n
(53-2)
y —
The integral in (53-2) can be evaluated by (52-6) in which the
prisms are added first in the z-direction or by (52-7) in which
the prisms are added first in the ^-direction.
It should be noted that formulas (52-6) and (52-7) give the
value of the area of the region R if the function f(x, y) = 1 ;
for the left member becomes
which is A. A can be evaluated by
dx dy or
j
dy
§54
MULTIPLE INTEGRALS
179
Example. Find the volume of the tetrahedron bounded by the
jC II 2
plane — h T + ~ = 1 and the coordinate planes (Fig. 44). Here,
a o c
—('--: -9-
If the prisms are summed first in the ^-direction, they will be summed
from x = 0 to the line a&, whose equation is
Therefore,
fb / X" 32/\|«(l
— c I ( x — H~ -- r ) V
Jo V 2a b /|o
,
dy
*
This result was obtained by using (52-6) for the evaluation of V, but
(52-7) could be used equally well.
64. Triple Integrals. The triple
integral is defined in a manner
entirely analogous to the definition
of the double integral. The
function
}(•£> y> z)
is to be continuous and single
valued over the region of space R x'
enclosed by the surface S. Let R
be subdivided into subregions ARlJk. If &Vl]k is the volume of
ARl]k, the triple integral of f(x, y, z) over R is defined by
FIG. 44.
(54-1) fRf(x,V,z)dVm lim
•'** n,m,p— i
i = nj = m,k «= p
by exactly the same argument as that used in Sec. 52.
In order to evaluate the triple integral, R is considered to be
subdivided by planes parallel to the three coordinate planes, the
180 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §5*
case of the rectangular parallelepiped being treated first. In this
case,
A IV = &x% by, A^.
By suitably arranging the terms of the sum
n,j
&;*, I**, ft,*.)
dy dz.
it can be shown, as in Sec. 52, that
(54-2) f f(x, y,z)dV = F f" F f(x, y, a
a/ ft »/ *0 %/ tf 0 %/ XO
By other arrangements of the terms of the sum, the triple integral
can be expressed by means of iterated integrals in which the order
of integration is any permutation of that given in (54-2).
If R is not a rectangular parallelepiped, the triple integral
over R will be evaluated by iterated integrals in which the limits
for the first two integrations will be functions instead of con-
stants. By extending the method of Sec. 53, it can be shown that
(54-3) fR f(x, y,z)dV = £' f™ fj^ f(x, y, z) dx dy dz.
Similarly, the triple integral can be evaluated by interchanging
the order of integration in the iterated integral and suitably
choosing the limits.
The expression (54-3), or the similar
expressions obtained by a different choice
of the order of integration, gives the
formula for the volume of R in case
•»vc f(x> y> *0 = 1- Therefore,
V =
dx dy dz.
Fm. 45. AlsQ? the formula (54.3) may b
sidered as giving the total mass of the volume V that has vari-
able density /(x, y, z).
Example. Let it be required to find the moment of inertia Ix of the
solid bounded by the cylinder #2 + t/2 = a2 and the planes 2 = 0 and
z = 6 about the s-axis (Fig. 45). Assume uniform density 0. The
function f(x, y, z) is the square of the distance of any point P(x, y, z)
from the z-axis. Therefore,
§54
Hence,
MULTIPLE INTEGRALS
181
p (a
Jo
2 + 62 - a2 sin2 0) cos2
_
4 i6 12
PROBLEMS
1. Evaluate
/ >, fir Ta(l— cos 0) , ,/>
W Jo Jo pdp^,
and describe the regions of integration in (a) and (6).
2. Verify that fR (x2 + y*) dy dx = fR (x* + y'2) dx dy, where the
region R is a triangle formed by the lines y = 0, y == #, and # = 1.
3. Evaluate and describe the regions of integration for
/ \ Ca fVa2
(«) JoJ«-«
4. Find the areas enclosed by the following pairs of curves:
(a) y = x, y = z2;
(6) y = 2~x, 2/2 = 2(2-*);
(c) y = 4 - x*, y = 4 - 2x-f
(d) y2 = 5 - x, y = x + 1;
(e) i/ =
a —
182 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §64
6. Find by double integration the volume of one of the wedges cut
off from the cylinder x2 + y2 = a2 by the planes z = 0 and z = x.
6. Find the volume of the solid bounded by the paraboloid
y't -f- %*• = 4$ and the plane x = 5.
7. Find the volume of the solid bounded by the plane 0 = 0, the
surface z = x2 + y2 + 2, and the cylinder x1 + y2 = 4.
8. Find the smaller of the areas bounded by y = 2 — x and
x2 + 2/2 = 4.
9. Find the volume bounded by the cylinders y = #2, ?/2 = z and
the planes 2 = 0 and 2 = 1.
10. Find the volume of the solid bounded by the cylinders
x2 + y2 = a2 ahd y2 + z2 = a2.
11. Find the coordinates of the center of gravity of the area enclosed
by y = 4 — z2 and y — 4 — 2x.
12. Find the moments of inertia about the x- and y- axes of the
smaller of the areas enclosed by y — a — x and x2 + y2 — a2.
13. Evaluate the following:
f<» f W-2/2 f \/a2-*2 j 7 7
Jojo Jo ctectedy;
14. Find by triple integration
(a) The volume in the first octant bounded by the coordinate
planes and the plane x + 2y + 82 — 4.
(6) The volume of one of the wedges cut off from the cylinder
x2 + y2 — a2 by the planes 2 = 0 and z = x.
(c) The volume enclosed by the cylinder x2 + y2 — 1 and the
planes 2 = 0 and 2 = 2 — #.
(d) The volume enclosed by the cylinders y2 = z and x2 + y2 = a2
and by the plane 2 = 0.
(e) The volume enclosed by the cylinders y2 + z2 = a2 and
r2 4- ^2 ^ ^
(/) The volume enclosed by y2 + 2z2 = 4z - 8, y2 + z2 = 4, and
x = 0.
(0) The volume in the first octant bounded by the coordinate
planes and x + 3y + 2z = 6.
(h) The volume enclosed by the cylinder x2 + y2 = 9 and the
planes 2 = 5 — x and 2 = 0.
(1) The volume of the cap cut off from y2 + z2 = 4# by the plane
2 = x.
15. Find the moments of inertia about the coordinate axes of the
solids in Prob. 14.
$55 MULTIPLE INTEGRALS 183
16. Find the coordinates of the center of gravity of each of the
volumes in Prob. 14.
17. Find by triple integration the moment of inertia of the volume
of a hemisphere about a diameter.
18. Find the coordinates of the center of gravity of the volume of the
solid in Prob. 17.
19. Find by triple integration the moment of inertia of the volume of
the c*. ne y1 + z2 = aV about its axis.
20. imd the moment of inertia of the cone in Prob. 19 about a
diameter of its base.
21. Find the volume in the first octant bounded by £ = # + !, z = 0,
y = 0, x = 2z, and z2 + i/2 = 4.
22. Find the coordinates of the center of gravity of the volume
bounded by z = 2(2 — x — y\ z = 0, and z = 4 — z2 — y2.
55. Jacobians. Change of Variable. If it is desired to make
a change of variable in a double or triple integral, the method is
not so simple as in the case of the simple integral. It is probably
already familiar to the reader that the element of area dA, which
is equal to dx dyin>' otangular coordinates, is not equal to dp dd in
polar coordinates. In order to obtain a general method for trans-
forming the element of area or the element of volume from one
set of coordinates to another, it is necessary to introduce the
definition of the Jacobian, or functional determinant.
Let u = u(xj y) and v = v(x, y) be two continuous functions
of the independent variables x and y, such that du/dx, du/dy,
j and dv/dy are also continuous in x and y. Then
(55-1)
du dv __ du dv
'dxdy ~~ ~dy^x
du dv
Jx dx
—
'dy dy
is called the Jacobian, or functional determinant, of u} v with
respect to x, y. It is usually denoted by
J I I Or -r~, r»
In the case of three variables, let u = u(x, y, z), v = v(x, y, z),
and w = w(x, y, z) be continuous together with their first partial
derivatives. The Jacobian, or functional determinant, of w, v, w
184 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §55
with respect to x, y, z is defined by
(55-2)
du dv dw
H)X 'dx ~dx
du to dw
dy lfy~dy
du dv dw
dz "dz ~dzi
The usual symbols for it are
or
d(u, vf w)
d(x, y, z) '
The Jacobian of any number of functions Ui, u^, • • • , un,
with respect to the variables xi, x%, - - - , xn, is defined by an
obvious extension of (55-1) and (55-2). It is denoted by
(Ui, U2, ' ' ' , Un\
\Xi, XZ, ' ' ' , Xn/
or
,Un)
The Jacobian is of great importance in mathematics.* It is
used here in connection with the change of variable in multiple
integrals. If it is desired to change the variable in fRf(x, y) dA
by making x = x(u, v) and y = y(u,v), the expressionf for dA in
terms of u and v is given by
(55-3)
dA =
(*LM\
\U,V/
du dv.
Thus, in transforming to polar coordinates by means of x =
p cos 0, y = p sin 8,
cos 6
sin 6
— p sin 6 p cos 0
= p cos2 0 + p sin2 6 = p.
Therefore,
dA = p dp d6,
a result that is already familiar from elementary calculus.
* Note that the Jacobian appeared in Sec. 41m connection with the differen-
tiation of implicit functions.
t See SOKOLNIKOFF, I. S., Advanced Calculus, Sec. 46.
§58 MULTIPLE INTEGRALS
It follows from (55-3) that
(55-4) f /(x, y) dA = f f f[x(u, v), y(u, v)]
JR J JR
The right-hand member of (55-4) can be written as
J fRF(u,v)dudv,
185
du dv.
where
F(u, v) = f[x(u, v), y(uy v}]
If it is desired to evaluate this double integral by means of an
iterated integral, the limits for u and v must be determined from
a consideration of the region R.
Similarly, if x = x(u, v, w), y = y(u, v, w)y and z = z(u, v, w),
then
u v w
du dv dw.
(55-5) dV = J ( -^-£'- ) du dv dw
and
(55-6) f f(x, y, z) dV
JR
s J J L f[x(u' Vj w}'
56. Spherical and Cylindrical Coordinates. Corresponding to
the system of polar coordinates in the plane, there are two
systems of space coordinates that are frequently used in prac-
tical problems. The first of these is the system of spherical, or
polar, coordinates. Let P(x, y, z) (Fig. 46) be any point whose
projection on the o^-plane is Q(x, y). Then the spherical
coordinates of P are p, <p, 0, in which p is the distance OP <p, is the
angle between OQ and the positive x-axis, and 6 is the angle
between OP and the positive 2-axis. Then, from Fig. 46, it is
seen that
x = OQ cos tf> = OP cos (90° — 0) cos <p = p sin 0 cos <?,
y = OQ sin <p = p sin 6 sin <p,
z = p cos 0.
The element of volume in spherical coordinates can be obtained,
by means of (55-5). Since
186 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
(x, V, g\ =
\P, V* OJ
sin 0 cos <p sin 0 sin <p cos 0
—p sin 0 sin ^> p sin 0 cos ^ 0
p cos 0 cos <f> p cos 0 sin ^ — p sin 0
s= — p2 sin 0,
it follows that
(56-1) dV = p2 sin
This element of volume is the volume of the solid bounded by the
Q(*,y)
FIG. 46.
two concentric spheres of radii p and p + dp, the two planes
through the z-axis that make
angles of <p and <p + d<p with
the £2-plane, and the two cones
of revolution whose common
axis is the z-axis and whose
vertical angles are 28 and
2(8 + dB).
The second space system cor-
responding to polar coordinates
in the plane is the system of
cylindrical coordinates. Any
point P(xt if, z), whose projection
on the £2/-plane is Q (Fig. 47), has
the cylindrical coordinates p, 8, z,
where 0 is the angle between OQ and the positive z-axis, p is the
distance OQ, and z is the distance QP. From Fig. 47, it is evident
FIG. 47.
§56 MULTIPLE INTEGRALS
that x = p cos 6, y = p sin 0, and 2 = 2. Since
187
(x y z\
[ x> y> z } —
\P> 0, zj
cos 0 sin 0 0
— p sin 0 p cos 6 0
0 0 1
it follows that
(56-2)
dV = pdp dO dz.
This element of volume is the volume of the solid bounded by the
two cylinders whose radii are p and p + dpy the two planes
through the z-axis that make angles 6 and 6 + d8 with the
zz-plane, and the two planes parallel to the :n/-plane at distances
z and z + dz.
Example 1. Find the re-coordinate of the center of gravity of the
solid of uniform density <r lying in the first octant and bounded by the
three coordinate planes and the sphere x1 + y2 + z* = a2.
Since
it is necessary to compute J R x dV. This integral can be calculated by
evaluating the iterated integral
fa /*Va2-z2 f Vo2-?/2
Jo Jo Jo
but it is easier to transform to spherical coordinates. Then,
7T 7T
C xdV = f2 f2 /°
IT IT
f~2 /*2 O4
= Jo Jo
sn cos
sn
Therefore,
oV /*|
= TeJ0
oV
_ _ 3a
7ra3/6 " 8'
Example 2. In the example of Sec. 54, find 7, by transforming the
integral into cylindrical coordinates. Then,
188 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §57
2V dv
*)p dz M dp
/V0
T
p3 sin 20 , 62p0\
— + -sv
(8.. + 46-).
57, Surface Integrals. Another important application of
multiple integrals occurs in the problem of defining the area of a
surface. Let z = f(x, y) be the equation of a surface 8 (Fig. 48).
Let 8' be a portion of this
surface bounded by a closed
curve F, and such that any
line parallel to the 2-axis cuts
Sr in only one point. If C is
the projection of T on the xy-
plane, let the region R, of
which C is the boundary, be
subdivided by lines parallel to
the axes into subregions ABt.
Through these subdividing
lines pass planes parallel to
the z-axis. These planes cut
from S' small regions Afi£ of area Aov Let AAt be the area of
AJ?t. Then, except for infinitesimals of higher order,
AA» = cos 7t Acrt,
where cos a», cos /3l; and cos 7» represent the direction cosines of
the normal to 8 at any point (xly y^ £t) of AS(. Since (see
Sec. 43)
dz
cos «<: cos &: cos 7* = -r-
ox
FIG. 48.
it follows that
cos 7» =
-1
§67 MULTIPLE INTEGRALS
Upon using the positive value for cos 7*,
189
Then,
is defined as the area of the surface S'. Since this limit is
«
dz
the value of <7 is given by
(.7.1)
-i
7
Similarly, by projecting S
on the other coordinate
planes, it can be shown that
= I sec a dA
jRi
= I sec /3 dA.
JR*.
The integral of a function
v(x, y, z) over the surface
z = f(x, y) can now be de-
fined by the equation
FIG. 49.
(57-2)
Js
, y, z) dcr
-Si
It is assumed that <p(x, y, z) is continuous and single valued for
all points of some region T that contains S.
Example. Find the area of that portion of the surface of the cylinder
x* + y2 = #2 which lies in the first octant between the planes z = 0 and
z = mx (Fig. 49).
' This surface can be projected on the zz-plane or on the t/z-plane but
not on the xy-pl&ne (since any perpendicular to the xy-pl&ne that meets
the surface at all will lie on the surface). The projection on the #z-plane
is the triangle GAB. Hence,
190 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §67
*~foAB "****.
But
rr— 0 = a(a2 -
V a — %
Therefore,
* "" ^"^ dz dx
C
I
amx(a2 — #2)-^ dx = a2w.
PROBLEMS
1. Find the coordinates of the center of gravity of the area bounded
by x^ + yM = a^, z = 0, and ?/ = 0.
2. Find the moment of inertia of the area of one loop of p2 = a2 sin 26
about an axis perpendicular to its plane at the pole.
3. (a) Find the expression for dA in terms of u and v, if x — u(l — v)
and y = uv.
(6) Find the expression for dV in terms of u, v, and w, if x = u(l — v),
y — uv(\ — w), and z = mw>.
4. Find the center of gravity of one of the wedges of uniform density
cut from the cylinder x2 + y2 = a2 by the planes 2 = mx and 2 = — mx.
6. Find the volume enclosed by the circular cylinder p ~ 2a cos 0,
the cone z = p, and the plane 2 = 0 (use cylindrical coordinates).
6. Find the center of gravity of the solid of uniform density bounded
by the four planes - 4- T + - = 1, # = 0, 2/ = 0, and 2 = 0.
01 0 C
7. Find the moment of inertia of the solid of uniform density
bounded by the cylinder x2 + y2 = a2 and the planes 2 = 0 and 2 = 6
about the 2-axis.
8. Find, by the method of Sec. 57, the area of the surface of the
sphere x2 + y2 + 22 = a2 that lies in the first octant.
9. Prove that
Hint: Write out the Jacobians, and multiply.
10. Prove that
w y/
where u = u(x, y), v = v(x, y), x = z(£, 17), and i/ = y(?, 17).
11. Find the surface of the sphere x2 + y2 + z* = a2 cut off by the
cylinder a;2 — ax + y* = 0.
§58
MULTIPLE INTEGRALS
191
12. Find the volume bounded by the cylinder and the sphere of
Prob. 11.
13. Find the surface of the cylinder x2 + y2 = a2 cut off by the
cylinder y* + z2 = a2.
14. Find the coordinates of the center of gravity of the portion of
the surface of the sphere cut off by the right-circular cone whose vertex
is at the center of the sphere.
15. Use cylindrical coordinates to find the moment of inertia of the
volume of a right-circular cylinder about its axis.
16. Find the moments of inertia of the volume of the ellipsoid
about its axes.
17. Kinetic energy T is defined as T — %Mv2, where M is the mass
and v is the velocity of a particle. If the body is rotating with a
constant angular velocity o>, show that
T =
where p is the density and / is the moment of inertia of the body about
the axis of rotation.
58. Green's Theorem in
Space. An important the-
orem that establishes the
connection between the in-
tegral over the volume and
the integral over the surface
enclosing the volume is given
next. This theorem has
wide applicability in numer-
ous physical problems* and
is frequently termed the di-
vergence theorem.
THEOREM. // P(x,y,z), FIG. 50.
Q(x, y, z), R(x, y, z) and
dP/dx, dQ/dy, dR/dz are continuous and single-valued functions in
a region T bounded by a closed surface S, then
It will be assumed that S (Fig. 50) is cut by any line parallel to
* See, in this connection, Sees. 125, 130, 131.
192 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §68
one of the coordinate axes in at most two points. If S is not
such a surface, then T is subdivided into regions each of which
satisfies this condition, and the extension to more general types
of regions is immediate.
A parallel to the z-axis may cut S in two points (xt, yr, zt) and
(zt, ?/t, zt), in which zl < zt. Let z = fi(x, y) be the equation
satisfied by (zt, t/t, zt) and z = /2(x, y) be the equation satisfied
by (zt, 2/t, Zt). Thus, S is divided into two parts, Si, whose equa-
tion is z = fi(x, y), and S2, whose equation is z = /2(#, y). Then,
R(x, y, z) cos 7 d<r,
taken over the exterior of S, is equal to
XR(XJ y, z) cos 7 do- + I /?(#, ?/, z) cos 7 rfo-,
'^ «/<Si
taken over the exteriors of the surfaces Si and S2. But, from
(57-2), these surface integrals are equal to double integrals taken
over the projection T' of T on the £i/-plane. Therefore,*
I R(x, y, z) cos 7 da = {#[>, 2/,/2(z, y)] - R[x, y,fi(x, y)]} dA
JS JT'
R(XJ y, z) dy dx
- dz
or
I R(x, y, z) cos 7 da = | -^ dV.
Similarly, it can be shown that
f f dP
I P(#, t/, z) cos a dcr = I — rfF
J-s Jr ^
fQCr,t/,z)cos/3d<r= ( $dV.
Js JT oy
* The negative sign appears in the right-hand member of the equation
because
COS 72 dffz ** ~" COS
where the subscripts refer to S* and Si.
and
§68 MULTIPLE INTEGRALS 193
Therefore,
(58-1) (P cos a + Q cos 0 + R cos 7) dr = + + d7.
Since cos ad<r = dy dz, cos /3 dv = dz dxy and cos y da = dx dy,
(58-1) can be written in the form
dR
(58-2) I I (P dy dz + Q dz dx + R dx dy)
m(dP ^dQ
\3x+^
The formula (58-2) bears the name of Green.*
Example. By transforming to a triple integral, evaluate
/ = J J (xs dy dz + x2y dz dx + #2z da; dy\
where >S is the surface bounded by z = 0, z = 6, and a;2 -f 2/2 = a2.
Calculating the right-hand member with the aid of (58-2) and mak-
ing use of the symmetry, one finds
/*a /• v/aT^lr* /*&
7 = 4 Jo Jo Jo (*** + ** + **) *>dv**
= 4-56 jTa a;2 V~a2 - x> dx
= %ira*b.
A direct calculation of the integral 7 may prove to be instructive. The
evaluation of the integral can be carried out by calculating the sum of
the integrals evaluated over the projections of the surface S on the
coordinate planes. Thus,
a ' /*\/a* — y2
.J_V^M-
which upon evaluation is seen to check with the result obtained above.
It should be noted that the angles a, j8, 7 are made by the exterior
normal with the positive direction of the coordinate axes.
* The names of Gauss and Ostrogradsky are also associated with this
theorem.
194 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §69
59. Symmetrical Form of Green's Theorem. One of the most
widely used formulas in the applications of analysis to a great
variety of problems is a form of Green's theorem obtained by
setting
„ dv ~ dv D dv
P = u—> Q = u—> R = u~
dx dy dz
in (58-1). The result of the substitution is
f fdv , &> a , dv \ j
I u I -r- cos a + — cos j8 + — cos 7 I ao*
js \dx dy dz /
C
==
Jr
T /^u
Jr \dx
5v 6u dv du ,„
'dx ^ ltydy + ~dz
But the direction cosines of the exterior normal n to the surface
are
dx n dy dz
cos a = -j-i cos jS = -r-) cos 7 = 3-7
dn an an
so that the foregoing integral reads
(59-1) f u ^ dff = I 11 V2i; dF
' Js dn JT
where
f (dudv_ , du dv_ du --i.y
Jr V^a; aa; ^ dy dy ^ dz dzj '
= <
dx2 + dy* + dz2'
Interchanging the roles of u and v in (59-1) and subtracting the
result from (59-1) give the desired formula
A reference to the conditions imposed upon P, Q, and R in
the theorem of Sec. 58 shows that, in order to ensure the validity
of this formula, it is sufficient to require the continuity of the
functions u and v and their first and second space derivatives
throughout a closed region T.
§59 MULTIPLE INTEGRALS 195
PROBLEMS
1. Evaluate, by using Green's theorem,
J Js
where S is the surface z2 + y2 + z2 = a2.
2. Show from geometrical considerations that the angle dd subtended
at the origin by an element ds of a plane curve C is
ds
dd = cos (ft, r) — >
where r is the radius vector of the curve, and (n, r) is the angle between
the radius vector and the normal to the curve. Hence, show that
r cos (n, r) ds r I dr
e = Jc — r
f 1 dr
= )c~rd^ds>
where the integral is a line integral along the curve C.
3. A solid angle is defined as the angle subtended at the vertex of a
cone. The area cut out from a unit sphere by the cone, with its vertex
at the center, is called the measure of the solid angle. The measure of
the solid angle is clearly equal to the area cut out by the cone from any
sphere concentric with the unit sphere divided by the square of the
radius of this sphere. In a manner analogous to that employed in
Prob. 2, show that the element of solid angle is
cos(n, r) da
&* = ~2 1
where the angle between the radius vector and the exterior normal to
the surface S is (n, r). Also, show that
cos (n, r) do" f 1 dr
x . _ • \ > / §
CO =
where the integral is extended over the surface S.
4. By transforming to a triple integral, evaluate
//,
dy dz + y3 dz dx + 23 dx dy\
where S is the spherical surface x2 + yz + z* = a2. Also, attempt to
calculate this integral directly.
5. Set v — 1 in Green's symmetrical formula, and assume that u
satisfies the equation^fc Laplace, V2u = 0. What is the value of
f -JT do- if S is an arDiwary closed surface?
196 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §59
6. The density of a square plate varies directly as the square of the
distance from one vertex. Find the center of gravity and the moment
of inertia of the plate about an axis perpendicular to the plate and
passing through the center of gravity.
7. Find the volume of a rectangular hole cut through a sphere if a
diameter of the sphere coincides with the axis of the hole.
8. Show that the attraction of a homogeneous sphere at a point
exterior to the sphere is the same as though all the mass of the sphere
were concentrated at the center of the sphere. Assume the inverse
square law of force.
9. The Newtonian potential V due to a body T at a point P is defined
by the equation V(P) = j T dm/r, where dm is the element of mass of
the body and r is the distance from the point P to the element of mass
dm. Show that the potential of a homogeneous spherical shell of
inner radius b and outer radius a is
V = 27r<7(a2 - 62), if r < 6,
and
4 a3 - &3
= ^TTO- -- - — > if r > a,
where cr is the density.
10. Find the Newtonian potential on the axis of a homogeneous
circular cylinder of radius a.
11. Show that the force of attraction of a right-circular cone upon a
point at its vertex is 2ir<rh(l — cos a), where h is the altitude of the
cone and 2a is the angle at the vertex.
12. Show that the force of attraction of a homogeneous right-circular
cylinder upon a point on its axis is
here h is altitude, a is radius, and R is the distance from the point to
one base of the cylinder.
13. Set up the integral representing the part of the surface of the
sphere #2 + yz + & — 100 intercepted by the planes x = 1 and x = 4.
14. Find the mass of a sphere whose density varies as the square of
the distance from the center.
15. Find the moment of inertia of the sphere in Prob. 14 about a
diameter.
CHAPTER VI
LINE INTEGRAL
The line integral, to be considered in this chapter, is as useful
in many theoretical and practical problems as the ordinary defi-
nite integral defined in Chap. V. The discussion of the line
integral will be followed by several
illustrations of its use in applied
mathematics.
60. Definition of Line Integral.
Let C be any continuous curve (Fig.
51), joining A(a,b) and B(c,d).
Let M(x, y) and N(x, y) be two
functions that are single- valued and
continuous functions of x and y for
all points of C. Choose n — 1 points
PI(XI, y^) on the curve C, which is
thus divided into n parts. Let
where XQ = a, yQ = 6, xn = c, yn = d.
Jxn
Fio. 51.
Let {» and rj t be defined by
< 7/t and form
t, 170 Ax,
The limit of this sum as n — > QQ and all Ax% —» 0 and
simultaneously is defined as a line integral along C.
Thus,
(60-1) lim V [M(^ i/O Ax, + tf (&, nO AyJ
n-^oo
[M(X, y) dx + N(x, y) dy].
Obviously, the value of this integral depends, in general, on the
particular choice of the curve C. If the equation of C is known
in one of the forms y = /(x), x = <p(y) or x = /i(0, t/ = /2(0, the
197
198 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §60
line integral may be reduced to a definite integral in one variable
by substitution, as is indicated in the following examples. How-
ever, it is frequently inconvenient to make this reduction, and
thus it is desirable to consider the properties and uses of (60-1).
Example 1. Let the points (0, 0) and (1, 1) be connected by the line
y = x. Let M(x, y) = x — y2 and N(xt y) = 2xy. Then the line
integral along y = x,
becomes, on substitution of y = #,
[(x - x2) dx + 2x2>dx] = (x + x2) dx
If (0, 0) and (1, 1) are connected by the parabola y = x2, I along
y = x2 is
[(x - x4) dx + 2x*(2x dx)] = o (x + 3z4) dx = %•
Example 2. Consider
M(x, y) = 2x2 + 4xy
and
N(xt y) = 2x2 - ?/,
with the curve y = x2 connecting the points (1,1) and (2, 4). Then
/*(2,4) /»2 /*!
J(u) (Mdx + Ndy)=Ji (2x2 + 4x-x2)dx+J} (2y - ?/2) dy = 13%.
Inasmuch as dy = 2o; do:, this integral can be written as
f2 (2x2 + 4x3) da: + f2 (2a;2 - o;4)2x dx = 13?^.
If the equation of the parabola in this example is written in a para-
metric form as
x = t,
y = t2, (l<t< 2),
then the integrand of the line integral can be expressed in terms of the
parameter t. Substituting for x, y, dx, and dy in terms of t gives
/*(2,4) /*2
J(u) (Mdx + N dy)} = Jt [(»' + 4«") + (24" - «<)2«] <tt
<2 + 8*3 - 2«6) dt =
§61 LINE INTEGRAL 199
The reader will readily verify that the value of this integral over a
rectilinear path C joining the points (1,1) and (2, 4) is also 13%J. In
fact, the value of this integral depends only on the end points and not
upon the curve joining them. The reason for this remarkable behavior
will appear in Sec. 63.
PROBLEMS
1. Find the value of I ' [\/y dx + (x — y) dy] along the follow-
»/(0,0)
ing curves:
(a) Straight line x — t, y = t.
(b) Parabola x = t2, y = t.
(c) Parabola x = t, y — t2.
(d) Cubical parabola x = t, y = t*.
2. Find the value of J^ [x2y dx + (x2 - y2) dy] along (a) y = 3s2,
(b) y = 3x.
3. Find the value of J (0[o) (x* dx + y2 dy) along the curves of
Prob. 1 above.
4. Find the value of J (0'0) [(x2 + y'2) dx — 2xy dy] along (a) y = x\
(V) x = y2; (c) y = x\
6. Find the value of J (Jo) (y s'n x ^x ~~ x cos y ^2/) a^ong y ~ x.
6. Find the value of J(-0 o> (x dy + y dx) along the upper half of
the circle x2 + y2 = a2.
7. Evaluate the integral of Prob. 6 over the path formed by the
lines x = — a, ty — a, x = a. What is the value of this integral if the
path is a straight line joining the points (—a, 0) and (a, 0)?
8. Find the value of /|J'J) (#2 dx + y2 dy) along the path given by
x = sin t, y = cos t.
9. Evaluate the integral of Prob. 8 if the path is a straight line join-
ing (0, 1) and (1, 0).
10. What is the value of the integral of Prob. 8 if the path is the
curve y = 1 — x2?
61. Area of a Closed Curve. Let C be a continuous closed
curve which nowhere crosses itself. The equation of such a
curve, in parametric form, can be given as
where the parameter t varies continuously from some value
200 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §61
t = to to t = t\ and the functions f\(t) and f2(t) are continuous
and single-valued in the interval to < t <> t\. Inasmuch as the
curve is assumed to be closed, the initial and the final points of
the curve coincide, so that
and
The statement that the curve C does not cut itself implies
that there is no other pair of values of the parameter t for which
and
A closed curve satisfying the condition stated above will be
called simple.
As t varies continuously from to to ti, the points (x, y) deter-
mined by (61-1) will trace out the curve C in a certain sense. If
C is described so that a man walking along the curve in the direc-
tion of the description has the
enclosed area always to his left,
the curve C is said to be described
in the positive direction^ and the
enclosed area will be considered
positive; but if C is described so
that the enclosed area is to the
right, then C is described in the
negative direction, and the area is
regarded as negative.
Consider at first a simple closed
curve C such that no line parallel
to one of the coordinate axes, say
the y-axis, intersects C in more
than two points. Let C be
, x = a2, y = 61, y = b2, which are
and J32, respectively. Clearly, C
FIG. 52.
bounded by the lines x
tangent to C at Ai, A*,
cannot be the graph of a single-valued function. Therefore, let
§61
LINE INTEGRAL
201
the equation of A\B\A^ be given by y\ = fi(x), and the equation
of AiBzAz by y* = /2(&), where /i (x) and/2(z) are single-valued
functions. Then the area enclosed by C (Fig. 52) is given by
(61-2)
or
(61-3)
= I 2/2 dz - fa2 2/1
Jai Jai
= - ] 2/2 da: - \
Jai Jai
A = - fcydx,
dx,
in which the last integral is to be taken around C in a counter-
clockwise direction.
Similarly, if x\ = <p\(y) is the equation of BiAiB^ and o:2 =
is the equation of
x-tdy — I *
Jb\
= I 2 x2 dy + C l
Jb\ Jbi
or
(61-4)
Again, the last integral is to be taken around C in a counter-
clockwise direction. It may be
noted that (61-3) and (61-4) both
require that the area be to the left
as C is described if the value of
A is to be positive.
By adding (61-3) and (61-4), a
new formula for A is obtained,
namely,
(61-5)
A ^
FIG. 53.
This formula gives a line-integral expression for A.
To illustrate the application of (61-5), the area between
(1) x2 = ty and (2) y2 = 4z (Fig. 53) will be determined.
202 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §62
Then,
A - o (~ydx + xdy} = H (-ydx + xdy)
* JC * J(l)
+ o (-
* J(2)
+ zc
24|0 24
16
3*
For convenience the first integral was expressed in terms of x,
whereas the second integral is simpler in terms of y.
The restriction that the curve C be such that no line parallel
to one of the coordinate axes cuts it in more than two points
can be removed if it is possible to draw a finite number of lines
connecting pairs of points on C, so that the area enclosed by
the curve is subdivided into regions each of which is of the type
considered in the foregoing. This extension is indicated in detail
in the following sect on.
PROBLEMS
1. Find, by using (61-5), the area of the ellipse x = a cos <pt
y = 6 sin <p.
2. Find, by using (61-5), the area between y2 = 9# and y = 3x.
3. Find, by using (61-5), the area of the hypocycloid of four cusps
x = a cos3 6, y = a sin3 6.
4. Find, by using (61-5), the area of the triangle formed by the line
x + y = a and the coordinate axes.
5. Find, by using (61-5), the area enclosed by the loop of the strophoid
62. Green's Theorem for the Plane. This remarkable theorem
establishes the connection between a line integral and a double
integral.
THEOREM. // M(x, y) and N(x, y), dM/dy and dN/dx are
continuous single-valued functions over a closed region R, bounded
by the curve C, then
§62 LINE INTEGRAL 203
The double integral is taken over the given region, and the curve C is
described in the positive direction.
The theorem will be proved first for a simple closed curve
of the typo considered in Sec. 61 (see Fig. 52).
Again, let y\ = fi(x) be the equation of AiBiA% and y% = /2(x)
be the equation of A\E^A^ Then,
. , , ,
-5— dx ay = I ax \ ~^-~ dy
; dy
[M(x, Vt) - M(x, y,}} dx
/*ai S*az
= - M(XJ 2/2) da? ~ I M(x,
Jai Jai
or
(62-1)
Similarly, if xi = <pi(y) is the equation of BiAiBi and
is the equation of B\A^.B^
na?
t
f ' tfdi,
J^z
or
(62-2) J J^ g dx dy = J^ JV^, y) <fo.
Therefore, if (62-2) is subtracted from (62-1),
(Sf ^ S) dx dy ~ ~ fc [M(x> & dx
It will be observed that setting M = — y and N ~ x gives the
formula (61-5).
Now, let the region have any continuous boundary curve C, so
long as it is possible to draw a finite number of lines that divide
the region into subregions each of the type considered in the
first part of this section; that is, the subregions must have
boundary curves that are cut by any parallel to one of the
coordinate axes in at most two points. Such a region R is shown
in Fig. 54.
204 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §62
By drawing the lines AiA2 and A3A4, the region R is divided
into three subregions R\, Rz, and Rs. The boundary curve of
each region is of the simple type. The positive direction of
each boundary curve is indicated by the arrows. The theorem
can be applied to each subregion separately. When the three
equations are added, the left-hand members add to give the
double integral over the entire region R. The right-hand mem-
bers give
- f (Mdx+Ndy)- f (Mdx + Ndy)- f (Mdx + Ndy),
JCi JCi JCs
where
Since each of the lines AiA2 and A3A4 is traversed once in each
direction, the line integrals that arise from them will cancel.
The remain ng line integrals,
taken over the arcs of C, add
to give the line integral over
C. Therefore,
nfdM _
\dy
--X
dx,
(Mdx + N dy)
holds for regions of the type
R.
Another type of region in
which an auxiliary line is in-
troduced is the region whose
boundary is formed by two or more distinct curves. Thus, if R
(Fig. 55) is the region between C\ and C2, the line A\A% is drawn
in order to make the total boundary
FIQ. 54.
a single curve.
integrals over A
integrals over C
The theorem can be applied, and the line
^A\ and A\A% will cancel, leaving only the line
and C2.
§62
LINE INTEGRAL
205
If the region R is such that any closed curve drawn in it can,
by a continuous deformation, be shrunk to a point without
crossing the boundary of the region, then the latter is called
simply connected. Thus, regions bounded by a circle, a rectangle,
or an ellipse are simply connected. The region R exterior to
C2 and interior to Ci (Fig. 55) is not simply connected because a
circle drawn within R and enclosing €2 cannot be shrunk to a
point without crossing C2. In ordinary parlance, regions that
have holes are not simply connected regions; they are called
FIG. 55.
FIG. 56.
multiply connected regions. The importance of this classification
will appear in the next two sections.
Example. Evaluate by using Green's theorem
where C is the closed path formed by y — x and i/3 = x* from (0, 0)
to (1, 1) (Fig. 56). Since M = x*y and N = y\
Then,
dN
.£«*»*+.•*> --//,(£-£)**
206 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §63
PROBLEMS
1. Find, by Green's theorem, the value of
jc (x2y dx + y dy)
along the closed curve C formed by y* — x and y — x between (0, 0)
and (1, 1).
2. Find, by Green's theorem, the value of
along the closed curve C formed by y* = #2 and y = x between (0, 0)
and (1, 1).
3. Use Green's theorem to find the value of
Jc [(*y - £2) dx + x*y dy]
along the closed curve C formed by y = 0, x = 1, and y = x.
4. Use Green's theorem to evaluate
along the closed path formed by y = 1, x = 4, and y — + -\/x.
5. Check the answers of the four preceding problems by evaluating
the line integrals directly.
63. Properties of Line Integrals. THEOREM 1. Let M and N
be two functions of x and y, such that M, N, dM/dy, and dN /dx are
continuous and single-valued at every point of a simply connected
region R. The necessary and sufficient condition that fc (M dx
+ N dy) = 0 around every closed curve C drawn in R is that
dM = dN
dy dx'
for every point of R.
Since
-]>«. + **>-//(£-£)**
where A is the region enclosed by C, it follows that
dM = dN
dy dx
LINE INTEGRAL
207
makes the double integral, and consequently the line integral,
have the value zero. Conversely, let J*c (Af dx + N dy) = 0
around every closed curve C drawn in R. Suppose that
dM _ «# ,0
dy dx *
at some point P of R. Since dM/dy and dN/dx are continuous
functions of x and y,
dM d#
is also a continuous function of x and y. Therefore, there must
exist some region S, about P, in which — has the same
sign as at P. Then,
ff (£-£)<•*'•*
J Js \ dy ox/
and hence f (M dx + N dy) T± 0
around the boundary of this re-
gion. This contradicts the hy-
pothesis that
fc (M dx + N dy) = 0
around every closed curve C drawn in R.
+x
FIG. 57.
It follows that
at all points of R.
Example 1. Let
M =
Then,
dM
dy
dN
dx
dM
dy
and
N =
dN
dx
M, N, dM/dy, and dN/dx are continuous and single-valued for all
points of the xy-plane except (0, 0). Hence, fc (M dx + N dy) = 0
around any closed curve C (Fig. 57) that does not enclose (0, 0). In
polar coordinates, obtained by the change of variables
208 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §63
x = p cos By y = p sin 6,
f ( r? g <fo + -TT— « *A = f de-
Jc \x2 -ft/2 a:2 -f ?/ V Jc
If C does not enclose the origin, 0 varies along C from its original value
00 back to 0o. Therefore, jc dO = 0. If Ci encloses the origin, 0 varies
along Ci from 00 to 00 + 2?r, so that J Cl dO = 2?r.
Example 2. Find, by Green's theorem, the value of
I = fc [(x* + xy) dx + (y* + x*) dy],
where C is the square formed by the lines y = ± 1 and x = ± 1. Since
dM dN
Note that the line integral has the value zero, but dM /dy ^ dN /dx.
This does not contradict Theorem 1.
THEOREM 2. Le^ M and ]V satisfy
the conditions of Theorem 1. 7"/ie
necessary and sufficient condition that
f(a'&) (^ ^ + ^ rfy) 6e independent
o/ i/ie citr^e connecting (a, b) and
A(a'b) (x, y) is that dM/dy = dN/dx at all
~x points of the region R. In this case
FIG. 58. the line integral is a function of the
end points only.
Suppose dM/dy = dN/dx. Let Ci and C2 (Fig. 58) be any two
curves from A to P, and let
7i = I (M dx + N dy) | 1
and
J2 = C(Mdx + N dy)
•/C2
be the values of the line integral from A to P along Ci and C2?
respectively. Then /i — 72 is the value of the integral around
the closed path formed by Ci and C2. By Theorem 1,
Jl - /2 = 0.
§63
LINE INTEGRAL
209
Therefore, Ii = /2, so that the line integral taken over any two
paths from A to P has the same value.
Conversely, suppose that J (M dx + N dy) is independent of
the path from A to P. Then, for any two curves Ci and C2,
/! = 72. It follows that / (M dx + N dy) = 0 for the closed
path formed by C\ and C2. Hence, by Theorem 1, dM/dy
= dN/dx.
Example. Consider
(22) /I + 7/2 1+z* \
m \x^dx — nr~vdyr
1,1) \ •" " /
Since dM/dy — 2y/x3 and dN/dx — 2y/xs and both functions are
continuous except at (0, 0), the line integral is independent of the
path so long as it does not enclose the origin. Choose y — 1 from
(1, 1) to (2, 1) and x = 2 from (2, 1) to (2, 2) as the path of integration.
Then,
25 ,__JL2__522__9
[ x2\i 81 8
THEOREM 3. Let M ami N satisfy the conditions of Theorem 1.
The necessary and sufficient con-
dition that there exist a function
F(x, y} such that dF/dx = M
and dF/dy = N is that dM/dy
— dN/dx at all points of the re-
gion R.
If dM/dy = dN/dx, Theorem
2 proves that
P(octy)
j (Mdx + N dy) FlG> 59<
is independent of the path. Therefore,
(63*1) f ("f (Mdx + N dy) = F(x, y},
a/ \u,o)
and this function F(x, y) depends only on the coordinates of the
end points of the path. Hence,
F(x + Ax, y) = f *6+A*'y) (M dx + N dy).
Let the path of integration be chosen as a curve C (Fig. 59)
from A to P and the straight line PP' from P(x, y) to P'(x + Ax,
210 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §63
y). Then,
F(x + Ax, y) =f^ (Mdx + N dy) + f^**'* (M dx + N dy)
or
(63-2) F(x + Ax, y) = F(x, y) + **** M(x, y} dx.
The second integral reduces to the simpler form given in (63-2)
since y is constant along PP', and therefore dy = 0. From (63-2),
,y)1
J
r j p+A* -i
= lim -r- I Jlf (a, y) rfx .
Ax_o LAx Jx v * ^ J
Application of the first mean-value theorem* gives
f xH~Ax M(x, y) dx = Ax M(f, »), (x ^ £ < x + Ax).
»/#
Therefore,
= lim - • Ax M(«, 2/) = lim Jlf ({, y).
OX Az-*0 LAX J Ax->0
Hence,
It can be proved similarly that
* It may be recalled that
>
f(x) dx = (6 — «)/(£), (a
is the first mean-value theorem for definite integrals. If ff(x)dx = F(x),
than /(a;) = /^'(x). From these relations, the mean-value theorem for
definite integrals can be transformed into
F(b) - F(a) = (6 - a)*"(0,
where a ^ ^ ^i 6, or
which is the mean-Value theorem of the differential calculus.
§63 LINE INTEGRAL 211
The function F is really a function of both end points. Multi-
plying dF/dx = M(x, y) by dx and dF/dy = N(x, y) by dy gives
dF dF
dF = ^dx + ^dy = M(x, y) dx + N(x, y) dy.
Thus, if
dM = dN
dy dx9
the integrand in Jc (M dx + N dy) is the exact differential
of the function F(x, y), which is determined by the formula
(63-1).
The most general expression for a function $(x, #), whose
total differential is d$ = M dx + N dy, is
*(*, y) = F(x, y) + C,
where C is an arbitrary constant. Indeed, since dF and d&
are equal,
d(F - $) = 0,
so that
F — 3> = const.
To prove the necessity of the condition of the theorem, note
that if there exists a function F(x, y) such that
then
OF dF
— = M(x, y) and — = N(x, y),
= a M , d*F = dN
dy dx ~ dy dx dy ~~ dx'
Since dM/dy and dN '/dx are both continuous, - — ^- and
are also continuous; hence,*
dx dy dy dx
Therefore,
dM = aAT
dy d«"
* See Sec. 46.
212 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §64
As a corollary to Theorem 3, one can state the following:
The necessary and sufficient condition that M(x, y) dx + N(x, y) dy
be an exact differential is that dM/dy = dN /dx.
PROBLEMS
1. Show that
.u [(x* + y^ dx + 2xy dy]
is independent of the path, and determine its value.
2. Test the following for independence of path:
(a) J (y cos x dx + sin x dy) ;
(b) f [(x* - y*) dx + 2xy dy];
(c) / [(x - ?/) dx + 2xy dy];
(d) f[(x?-y*)dx-2(x-l)ydy].
3. Show that
I ' j\ — L ' \s dx + n . — r« % is independent of
•J(0,o) L \i ~t~ £>> l-L T" 3J</ J
the path, and find its value.
4. Show that the line integral
-ydx xdy
evaluated along a square 2 units on the side and with center at the origin
has the value 2ir. Give the reason for failure of this integral to vanish
along this closed path.
5. Find the values of the following line integrals:
(y cos x dx + sin x dy) ;
,
(c) [(x + l)dx + (y + 1) dy].
64. Multiply Connected Regions. It was shown that the
necessary and sufficient condition for the vanishing of the line
integral Jc [M(x, y) dx + N(x, y) dy] around the closed path C
is the equality of dM/dy and dN '/dx at every point of the region
enclosed by C. It was assumed that C was drawn in a simply
connected region R and that the functions M(x, y) and N(xr y\
§64 LINE INTEGRAL 213
together with their first partial derivatives, were continuous on
and in the interior of C. The latter condition was imposed in
order to ensure the integrability of the functions involved. The
reason for imposing the restriction on the connectivity of the
region essentially lies in the type of regions permitted by Green's
theorem.
Thus, consider a region R containing one hole (Fig. 60). The
region R will be assumed to consist of the exterior of C2 and the
interior of C\. Let a closed contour C be drawn, which lies
entirely in R and encloses C%.
Now, even though the functions
M(x, y) and N(x, y) together with
their derivatives may be continu-
ous in R, the integral
Jc [M(x, y) dx + N(x, y) dy]
may not vanish. For let K be
, , . *i«. 60.
any other closed curve lying in R
and enclosing C^ and suppose that the points A and B of K
and C are joined by a straight line AB. Consider the integrals
JAPA JAB JBQB JBA'
where the subscripts on the integrals indicate the direction of
integration along the curves K, C, and along the straight line AB,
as is indicated in Fig. 60. Since the path AB is traversed twice,
in opposite directions, the second and the last of the integrals
above will annul each other, so that there will remain only the
integral along K, traversed in the counterclockwise direction,
and the integral along C, in the clockwise direction. Now, if M
and N satisfy the conditions of Theorem 1, Sec. 63, then
f (Mdx+Ndy) + f (M dx + N dy) = 0,
JGK JQC
where the arrows on the circles indicate the direction of integra-
tion. Thus,
(64-1) f (M dx + N dy) = f (M dx + N dy),
J&K JQC
both integrals being taken in the counterclockwise direction.
214 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §64
The important statement embodied in (64-1) is that the
magnitude of the line integral evaluated over a closed path in R,
surrounding the hole, has the same constant value whatever be
the path enclosing C%. This value need not be zero, as is seen
from a simple example already
mentioned in Sec. 63. Thus,
let the region R consist of the
exterior of the circle of radius
unity and with center at the ori-
gin and of the interior of a con-
centric circle of radius 3 (Fig. 61).
-y
The functions M =
and
N =
x2 + y2
and their deriva-
FIG.
x* + y*
tivcs, obviously satisfy the con-
ditions of continuity in R and on Ci and C2. Also, dM/dy
= dN/dx. But
X
— y dx
\dyj,
where C is the circle
gives
f
Jo
x = a cos 0,
y = a sin 6,
a2 sin2 £ + a2 cos2
(1 < a < 3),
dO = 2ir.
The function F(x, y), of which M(x, y) dx + N(x, y) dy is
an exact differential, is F(x, y) = tan"1 -> which is a multiple-
valued function.
The function
F(x, y) =
[M(x, y) dx + N(x, y) dy],
where M and N satisfy the conditions of Theorem 1, Sec. 63, will
be single-valued if the region R is simply connected (as is required
in Theorem 1) but not necessarily so if the region is multiply
connected.
§65 LINE INTEGRAL 215
65. Line Integrals in Space.1 The line integral over a space
curve C is defined in a way entirely analogous to that described in
Sec. 63.
Let C be a continuous space curve joining the points A and J5,
and let P(x, y, z),,Q(x, y, z), and R(x, y, z) be three continuous,
single-valued functions of the variables x, y, z. Divide the
curve C into n arcs Ast, (i = 1, 2, • * • , n), whose projections
on the coordinate axes are &X{, Ayt, Azr, and form the sum
n
PCfc, i?., fO Az, + Q(&, *, fO At/, + B(ft, 77t, fO AsJ,
where (&, r;t, ft) is a point chosen at random on the arc Ast.
The limit of this sum as n increases indefinitely in such a way
that all As* — > 0 is called the line integral of P dx + Q dy + R dz,
taken along C between the points A and B., It is denoted by
the symbol
(65-1) fc [P(x, y, z) dx + Qfo y, z) dy + R(x, y, z) dz].
The conditions imposed upon the functions P, Q, and R are
sufficient to ensure the existence of the limit, provided that the
curve C is suitably restricted.
If the equation of the space curve C is given in parametric
form as
(65-2) ^ y
where /i (0, /2(0, and /8(0 possess continuous derivatives in the
interval to < t < fa, the line integral (65-1) can be expressed
as a definite integral
where P, Q, and R are expressed in terms of t with the aid of (65-2).
It is possible to derive three theorems analogous to those given
in Sec. 63 for line integrals in the plane. They are as follows:
THEOREM 1. Let the region of space considered be one in which
P(x, y, z), Q(x, y, z\ and fl(x, ?/, z) and their partial derivatives are
continuous and single-valued functions of x, y, and z. Then the
necessary and sufficient condition that
216 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §66
/ (P dx + Q dy + R dz) = 0
around every closed curve in the region is that
dP = dQ dQ^dR dR = dP
dy dxJ dz dy' dx dz'
for every point of the region.
THEOREM 2. Let the functions considered satisfy the conditions
of Theorem 1. Then the necessary and sufficient condition that
f (7f (Pdx + Qdy + R dz)
J (a,b,c)
be independent of the path from (a, fr, c) to (x, y, z) is that
dP dQ dQ dR dR dP
dy dx dz dy dx dz
for every point of the region.
THEOREM 3. Let the functions P ', Q> and R satisfy the conditions
of Theorem 1. Then, the necessary and sufficient condition that
there exist a function F(x, y, z) such that
is that
•\T/1 f\pj f\Tj\
dx ~~ ' dy ~" ^ ^ ""
aP^dQ dQ = dR dR = dP
di/ " 5o:; dz " dy' dx ~~ ^2;'
for every point of the region. The function F(x, y, z) is given by the
formula
F(x, y, z) = (Pdx + Qdy + R dz).
COROLLARY. The necessary and sufficient condition that
Pdx + Qdy + Rdz
be an exact differential of some function $(x, y, z) is that
dP^dQ dQ^dR dR = <>P_
dy dx' dz dy' dx dz'
for every point of the region. The function 3>(x, y, z) is determined
from the formula
*(x, y, z) = f(7f (Pdx + Qdy + R dz) + const.
J(a,o,c)
LINE INTEGRAL
217
t+1
FIG. 62.
These results are of particular importance in hydrodynamics and
the theory of electromagnetism. The vector derivation and
interpretation of these results are given in Chap. IX on Vector
Analysis.
66. Illustrations of the Application of the Line Integrals.
1. Work. It will be assumed that a force F(x, y) acts at every
point of the xi/-plane (Fig. 62). This force varies from point to
point in magnitude and direction.
An example of such conditions is the
case of an electric field of force. The
problem is to determine the work
done on a particle moving from the
point A (a, b) to the point B(c, d)
along some curve C. Divide the arc
AB of C into n segments by the -^
points Pi, P^ - • • , Pn-i, and let
Ast = PtPt+i. Then the force acting
at Pt is F(x%, 7/r). Let it be directed along the line PtS, and let
Ptjf be the tangent to C at Pt, making an angle 0t with Pt>S.
The component of F(x^ yl) along PiT is F cos 0t and the ele-
ment of work done on the particle in moving through the distance
Ast is approximately F(xly yl) cos 0t Ast. The smaller Ast, the
better this approximation will be. Therefore, the work done in
moving the particle from A to B along C is
W = lim J? F(xlf 7/t) cos 0r Ast = f F(x, y} cos 0 ds.
If a is the inclination of PtS and ft is the inclination of PtT,
then 0 = a — /3 and cos 0 = cos a cos 0 + sin a sin 0, so that
(66-1) W = f F(x, T/) (cos a cos ft + sin a sin ft} ds.
From the definition of a, it is evident that
F cos a = ^-component of F = X,
F sin a = ^-component of F = F.
Moreover, since cte/ds = cos /3 and dy/ds = sin £,
cos ft ds ^ dx and sin ft ds ~ dy.
218 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §66
Therefore, (66-1) becomes
which is a line integral of the form (60-1).
If C is a space curve, then an argument in every respect similar
to the foregoing shows that the work done in producing a dis-
placement along a curve C in a field of force where the components
along the coordinate axes are Z, 7, and Z is
W = fc (X dx + Y dy + Z dz).
To illustrate the use of this formula, the work done in displac-
ing a particle of mass m along some curve C, joining the points
FIG. 63.
A and B, will be calculated. It will be assumed that the particle
is moving under the Newtonian law of attraction
where k is the gravitational constant and r is the distance from
the center of attraction 0 (containing a unit mass) to a position
of the particle (Fig. 63).
The component of force in the direction of the positive #-axis
is
r-. « / \ l\i!H> JL>
F cos (x, r) = - -^ • -•
Similarly,
and
Z = -
km z
§66 LINE INTEGRAL 219
The work done in displacing the particle from 'A to B is
C
= -
i) A
B
W
A '
But
,—— — — — x dx + y dy + zdz
r = vz + y2 + z2 and rfr = - y * ' --
Therefore,
= — fcm I -5- = few - >
J^ r2 Lr_M
which depends only on the coordinates of the points A and B
and not on the path C. Denoting the distances from 0 to A
and B by r\ and r2, respectively, gives
l l\
)•
r2 ri/
W = km
The quantity & = Arn/r is known as the gravitational potential
of the mass m. It is easily checked that
-. -r— ; : — — — ; / — ~ — ;
dx dy . dz
so that the partial derivatives of the potential function $ give
the components of force along the* coordinate axes. Moreover,
the directional derivative of 4> in any direction s is
cM> __ d$ dx d$ dy , ^^> dz
ds dx ds dy ds dz ds
= X cos (x, s) + Y cos (y, s) + Z cos (z, s)
where F8 is the component of force in the direction s.
A conservative field of force is defined as a field of force in
which the work done in producing a displacement between two
fixed points is independent of the path. It is clear that in a
conservative field the integral
fa
'dz)
along every closed path is zero, so that the integrand is an exact
differential.
220 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §66
2. Flow of a Liquid. Let C be a curve on a plane surface across
which a liquid is flowing. The xi/-plane will be chosen to coincide
with the surface. The lines of flow are indicated in Fig. 64 by the
curved arrows. It will be assumed that the flow of the liquid
takes place in planes parallel to the xy-plane and that the depth
of the liquid is unity. The problem is to determine the amount
of liquid that flows across C in a unit of time.
If vt is the velocity of the liquid and OL% is the inclination of the
tangent to the line of flow at Pz, then vx\l = vl cos <xt is the ^-com-
ponent of vt and vv\t = vt sin at is
the ?y-componcnt of v%. Let Ast
denote the segment PtPt+i of C.
A particle at Pl will move in time
At to P', while a particle at Pl+i will
move to P;+i. Therefore, the
amount of liquid crossing PJPt+i in
time At is equal to the volume of
the cylinder whose altitude is unity
and whose base is P f l+\P(+iP(.
Aside from infinitesimals of higher
order, this volume is
AFt - PJP( • P*Pt+1 sin 0t,
FIG. 64.
in which 0t denotes the angle between PJ*( and PJPt+i. But
i = Ast and, except for infinitesimals of higher order,
^ = vl At. Therefore, AVl = vl At - Asr sin 0>. The volume
of liquid crossing C in a unit of time is
n
V = lim V vt sin 0t Ast.
n-*« ^
If T< denotes the inclination of the tangent to C at Pt, then
r» = 0t + a». Therefore,
Vt sin 0i As% = yt(sin rt cos «t — cos rt sin at) Ast
== yt cos al sin rt Ast — vr sin at cos rt Ast
Hence,
(66-2)
V = fc ( -vy dx + vx dy)
is the line integral which gives the amount of liquid that crosses
C in a unit of time.
§66 LINE INTEGRAL 221
If the contour C is a closed one and the liquid is incompressible,
then the net amount of liquid crossing C is zero, since as much
liquid enters the region as leaves it. This is on the assumption,
of course, that the interior of C contains no sources or sinks.
Thus, a steady flow of incompressible liquid is characterized by
the equation
/*
— vy dx + vxdy) = 0,
over any closed contour not containing sources or sinks. This
implies that (see Sec. 63)
(66-3) - %* = J-",
dy dx
which is an important equation of hydrodynamics known as the
equation of continuity. Moreover, from Theorem 3, Sec. 63, it is
known that there exists a function ^ such that
This function ^ is called the stream function, and it has a simple
physical meaning, for
(-vydx +vx dy)
represents the amount of liquid crossing, per unit time, any curve
joining (a, 6) with (x, y).
The function defined by the integral
(66-5) *(x, y) = (vx dx + vy dy)
is called the velocity potential. It is readily shown that
(66-6) §£-„. and fy = V
Upon comparing (66-4) with (66-6), it is seen that
d$ _ <W , ^5__^
dx " dy dy ~~ dx'
These are the celebrated Cauchy-Riemann differential equations.
If the integral (66-2) around a closed curve C does not vanish,
then the region bounded by C may contain sources (if V is
222 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §66
positive) or sinks (if V is negative). The presence of sources
or sinks is characterized by the singularities of the function ^,
that is, those points for which ^ is not continuous or where its
derivatives may cease to be continuous.*
The foregoing discussion is readily generalized to a steady
flow of liquids in space. Instead of the integral (66-5), one will
have
$(s, y, z) = f(*'"'* (vx dx + vydy + vz dz),
and if the integral is independent of the path C, the equations
corresponding to (66-3) are
c^x__c^, = n ^£_^V = A dVx dVz = 0
dy dx ' dy dz ' dz dx
In such a case the integrand is an exact differential, and the
velocity potential $(#, yy z) gives
3. Thermodynamics. A thermodynamical state of any sub-
stance is found to be characterized by the following physical
quantities: (1) pressure p, (2) volume v, and (3) absolute tempera-
ture T. The pressure, volume, and temperature are connected
by the equation
(66-7) F(p, v, T) = 0,
so that any two of the three quantities p, v, and T will suffice to
determine completely the state of the substance.
In the case of an ideal gas enclosed in a receptacle, Eq. (66-7)
has the form
pv - RT = 0,
where R is a constant. Let p and v be chosen to determine the
state of the gas, and consider p and v as the coordinates of a point
P in the py-plane. As the state of the gas changes, the point P,
which characterizes the state, will describe some curve C in the
py-plane. If the process is cyclic, so that the substance returns
to its original state, then the curve C will be a closed one.
It is important to know the amount Q of heat lost or absorbed
by the gas while the gas in the receptacle (for example, steam in
* See in this connection Sec. 64.
§66 LINE INTEGRAL 223
an engine cylinder) changes its state. Let Ap, Ay, and A!F be
small changes in the pressure, volume, and temperature, respec-
tively. Now if any two of these quantities, say p and vy do not
change, then the amount of heat supplied is nearly proportional
to the change in the remaining quantity. If all three quantities
change, then the total change AQ in the amount of heat supplied
is approximately equal to the sum of the quantities AQi, A$2,
and AQ3, due to changes Ap, Ay, and AT, respectively.*
Thus,
AQ = AQ1 + AQ2 + AQ3
= d Ap + c2 Ay + c3 AT7,
where ci, c%, and c3 are constants of proportionality. Then, the
total amount of heat supplied in the process is given by the
equation
(66-8) Q = (ci dp + c2 dv + c3 dT).
Solving (66-7) for T in terms of p and v gives T = /(p, y), so that
dT = ^-dp + ^dv.
dp ^ dv
If this expression is substituted in (66-8), one obtains
(66-9) Q = j'c [(Cl + c» |) dp + (c, + ca g) dv],
where the integration is performed over the curve C in the
py-plane, which is called the pv diagram.
Consider the state of the gas in the cylinder of a steam engine,
and let the piston be displaced through a distance As. Then,
if the area of the piston is A, the work AW performed by the
piston is given by
ATF = pA As = p Aw,
and the total work W performed during one cycle is
W = fcpdv.
It follows from (61-4) that this is precisely equal to the area of
the pv diagram.
18 This principle is called the principle of superposition of effects.
224 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §66
In deriving (66-9), it was assumed that p and v were the
independent variables, and it followed, upon making use of (66-7),
that the increment of heat is given by*
dQ = (ci + C3~-} dp + f €2 + cz~\dv
ss P(p, v) dp + V(p, v) dv,
where P and V are known functions of p and v. The expression
for dQ is not, in general, an exact differential (that is, dP/dv
?£ dV/dp), for the line integral (66-9) need not vanish for a
cyclic process. However, it is possible to show that the difference
between dQ and the work p dv is an-oxact differential, namely,
(66-10) dU ss dQ - pdv,
where the function U is called the internal energy of the gas.
It is also possible to showf that the ratio of dQ to the absolute
temperature, namely,
(66-11) dS^^Q
is likewise an exact differential. The function S is called the
entropy, and it plays a fundamental role in all investigations in
thermodynamics.
The formulas (66-10) and (66-11) can be used to show that
for an isothermal process (that is, when dT = 0)
dQ = p dv,
so that all the heat absorbed by the gas goes into the performance
of the work p dv. If the process is adiabatic (that is, such that
there is no gain or loss of heat), then dQ = 0 and, therefore,
dS = 0. It follows that the entropy S is constant during such a
process.
* By making use of T and v, or T and p, as the independent variables, it is
possible to write down two other important expressions for dQ.
f These assertions follow from the first and second laws of thermo-
dynamics.
CHAPTER VII
ORDINARY DIFFERENTIAL EQUATIONS
67. Preliminary Remarks. The great usefulness of mathe-
matics in the natural sciences derives from the fact that it is
possible to formulate many laws governing natural phenomena
with the aid of the unambiguous language of mathematics.
Some of the natural laws, for example those dealing with the
rates of change, are best expressed by means of equations involv-
ing derivatives or differentials.
Any function containing variables and their derivatives (or
differentials) is called a differential expression, and every equation
involving differential expressions is called a differential equation.
Differential equations are divided into two classes, ordinary and
partial. The former contain only one independent variable and
derivatives with respect to it. The latter contain more than one
independent variable.
The order of the highest derivative contained in a differential
equation is called the order of the differential equation. Thus,
dx*
is an ordinary differential equation of order 2, and
is a partial differential equation of order 3.
When a differential equation can be expressed as a polynomial
in all the derivatives involved, the exponent of the highest
derivative is called the degree of the equation. In the foregoing
examples the degree of the ordinary equation is £ and that of the
partial differential equation is 2. It should be observed that the
degree of
dx* + \G
is 2, when this equation is rationalized.
225
226 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §67
If an ordinary differential equation is of the first degree in
the dependent variable and all its derivatives, it is called a
linear differential equation. The general form for a linear
differential equation of the nth order is
where the PI(X) and f(x) are functions of x only.
An explicit function y = f(x), or an equation <p(x, y) = 0
which defines y as an implicit function of .r, is said to be a solution
of the differential equation
(67-1) F[x, y, yf, y", • - - , y^] = 0,
provided that, whenever the values of y, y', y", - - • , y(n)
are substituted in the left-hand member of (67-1), the latter
vanishes identically.
For example,
(67-2) ^ + y cos x = 0
has a solution
y = e~*mx, or log y + sin x = 0,
because the substitution of y and y' calculated from either one
of these expressions reduces (67-2) to an identity 0 = 0. Thus,
differentiation of the second equation gives - -~ + cos x = 0,
y ax
so that yf = — y cos #, and substitution in (67-2) gives 0 = 0.
The graph of a solution of an ordinary differential equation is
called an integral curve of the equation.
PROBLEM
Classify the following differential equations, and determine their
orders and degrees:
ox* dx oy oy
(c) -^ + sin y + x = 0;
ORDINARY DIFFERENTIAL EQUATIONS
227
(e) y" - vT=T« y' + by = 0;
•*' dt2 ~~ dx2'
(g) y" + x*y' + xy = sin x;
68. Remarks on Solutions.
the first order,
(68-1) ^ -
Consider a differential equation of
where f(x, y} is a single-valued and continuous function of the
variables x and y. If a point (JQ, ?/Q) is chosen in the xy-
and its coordinates are substituted in (68-1), then
= /u 2/0)
determines a direction associated with the point (#0, 2/o), since
dy/dx can be interpreted geometrically as the slope of an integral
curve. If a second point (?i, yi) is chosen and its coordinates
are substituted in (68-1), a direction is
associated with (xi, yi). Continuing in
this way, it is possible to find a direction
associated with every point of the plane
for which f(x, y) is defined. Now, sup-
pose that a point (#0, 2/o) is chosen in the
plane (Fig. 65) and the direction associated
with this point is determined. Let (#1, 7/1) be a point very near
to (XQ, yo) and in the direction specified by
FIG. G5.
Then,
dy
dx
dy
dx
determines a new direction. Upon proceeding a short distance
in this new direction, a third direction given by
228 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §68
is determined by the selection of a point (x2, 1/2) which is close to
(xi, yi). If this process is continued, there will be built up a
curve made up of short straight-line segments. If the points
(x0, 2/o), (xi, t/i), (z2, 2/2), ' ' ' , (xn, 2/n) are chosen very close
together, it becomes intuitively clear that this series of straight-
line segments approximates a smooth curve associated with the
initial point (XQ) t/0). Evidently, the equation of this curve will
be a solution of the differential equation (68-1), for the slope of
the curve is
In general, a different choice of (XQ, 2/o) will lead to a different
integral curve and thus to a different solution of (68-1).
The foregoing discussion forms the basis of one method of
graphical solution of differential equations of the first order.
Another important method of approximate solution of differential
equations is the method of infinite series, which is outlined next.
Let it be supposed that the function /(#, y) in (68-1) can be
expanded in Taylor's series about the point (x0, 2/0) ; then the
solution of (68-1) can be obtained in the form of a power scries in
x — XQ. Indeed, denote the solution of (68-1) by
(68-2) y = F(x).
Then, if the integral curve defined by (68-2) is to pass through
(XQ, 2/0), it is necessary that
y = F(xQ) = 2/0.
Substituting the coordinates of (XQ, 2/0) in (68-1) gives
•J~ = /O&O, 2/o) = F'(XQ).
Differentiating (68-1) yields
d*y = df(x, y) , df(x, y) dy^
dx2 dx dy dx
so that the value of the second derivative of (68-2) at XQ is
dy
§68 ORDINARY DIFFERENTIAL EQUATIONS 229
The formula (68-3) can be used to calculate d3y/dx3, and its
value at the point (#0, 2/0) can be obtained, for the values of the
first and second derivatives of F(x) at x = XQ are already known.
In this manner, one can attempt to find the solution of (68-1)
in the form of the series
y = F(*0) + F'(x0)(x - xa) + - (x - z0)2 + • • • .
In essence, this method of solution is the same as the method of
undetermined coefficients that is discussed in Sec. 98. Another
important method, due to the French mathematician E. Picard,
is discussed in Sec. 103.
Next consider a family of curves
(68-4) y = x2 + c,
where c is an arbitrary constant. Differentiation of (68-4) gives
-
-7— — AX.
dx '
which is the differential equation of the family of curves (68-4),
and which is free from arbitrary constants. If the given func-
tional relation contains two arbitrary constants, as, for example,
y = ci sin"1 x + c2,
then it is possible to eliminate these constants c\ and Cz by
two differentiations. The first differentiation gives
Solving for c\ yields
and differentiation of this equation gives
dy = Q
dx2 1 — x2 dx
This is a differential equation of the second order, and clearly it
has y = ci sin"1 x + 03 as a solution. It should be observed that
two differentiations were necessary in order to eliminate two
arbitrary constants.
230 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §68
In general, if f(x9 y, ci, c2, • • • , cn) = 0 is a functional
relation involving n arbitrary constants and defining y as a
function of x, then n successive differentiations will produce n
equations involving derivatives up to and including those of the
nth order. These n equations together with the given equation
/Or, t/, Ci, c2, • • • , cn) = 0 can be used to eliminate the n con-
stants ci, C2, • • • , cn, and the result will be a differential equation
of the nth order whose solution is f(x, y, Ci, c2, • • • , cn) = 0.
It can be shown that, in general, a differential equation of the
nth order has a solution which contains n arbitrary constants.
Moreover, no solution of a differential equation of the nth order
can contain more than n arbitrary constants. A solution that
contains n arbitrary constants is called the general solution of the
differential equation.
The foregoing discussion does not prove these facts. It
merely suggests that a functional relation containing n arbitrary
constants leads to a differential equation of order n. For the
proof of this theorem and its converse, any advanced treatise on
differential equations* can be consulted.
Any solution that is obtained from the general solution by
specifying the values of the arbitrary constants is called a par-
ticular solution. Particular solutions arc usually the ones that
are of interest in applications of differential equations. It
should be remarked, however, that some differential equations
possess solutions which cannot be obtained from the general
solution by specifying the values of the arbitrary constants.
Some examples illustrating the existence of such solutions are
given in Sec. 83.
PROBLEMS
Find the differential equations of the following families of curves :
1. x* + cx + y = c2.
2. ci sin x + c2 cos x = y.
3. Cix + c&x + C&-* = y.
4. ce* - xy + e~x = 0.
6. (x - Cl)2 + (y - c2)2 = 1.
6. y = c\ex sin x + c2e* cos x.
7. c2x + cy + 1 = 0.
8. cfy + ciy + c2 = 0.
9. y = Cix* + c&2 + c&.
* See INCE, E. L. Ordinary Differential Equations.
§69 ORDINARY DIFFERENTIAL EQUATIONS 231
10. yz - 4cx = 0.
11. y = Ci62aj + c2e3* + #.
69. Newtonian Laws. In order to illustrate the prominence
of the subject of differential equations in a study of various
phenomena, the next four sections are primarily concerned with
the task of setting up the differential equations from the basic
physical principles. A systematic treatment of the problem of
solving various typos of differential equations frequently occur-
ring in practice will be given in the subsequent sections.
The formulation of the basic principles from which many
differential equations arise rests on the following fundamental
laws of dynamics, which were enunciated by Sir Isaac Newton.
1. Every particle persists in its state of rest or moves in a straight
line with constant speed unless it is compelled by some force to
change that state.
2. The rate of change of momentum of a particle is proportional
to the force acting on it and is in the direction of the force.
3. Action and reaction are equal and opposite.
The first law merely states that any change of velocity of a
particle (that is, acceleration) is the result of some external force.
The second law postulates that the resultant force / acting on a
particle is proportional to the product of the mass m by its accel-
eration a; for momentum is defined as the product of mass m and
velocity v, and the rate of change of momentum is
d , ^ dv
_ (mv) = m -rr = ma.
at at
Thus,
ma = kf,
where k is the proportionality constant, which can be made equal
to unity by a proper choice of units.
Obviously, the second law includes the first; for if the force
acting on a particle is zero, then its acceleration is zero and the
particle must either remain at rest or move with constant
velocity.
The third law asserts that, if two particles exert forces on each
other, then the force exerted by the first on the second is equal to
the force exerted by the second on the first. This law can be
used to define the mass of a body.
232 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §69
Frequent use of these laws will be made in the following pages.
There is one more law, formulated by Newton, that will be found
of cardinal importance in this study. It is the law of gravita-
tion. Newton was led to it by his attempts to explain the
motions of the planets. This law states that two bodies attract
each other with a force proportional to the product of their
masses and inversely proportional to the square of the distance
between them, the distance being large compared with the
dimensions of the bodies. If the force of attraction is denoted
by F, the masses of the two bodies by mi and m%, and the distance
between them by r, then
(69-1) F = ^p,
where K is the proportionality constant, called the gravitational
constant. In the c.g.s. system the value of K is 6.664 X 10~8.
The three fundamental principles formulated by Newton in
reality form the postulates of dynamics and furnish a definition
of force, and the law of gravitation permits one to compare
masses with the aid of the beam balance.
The law of attraction (69-1) assumes a simpler form in the case
of a small body of mass m falling to the earth from heights that
are not too great. It can be established that a sphere attracts a
particle at an external point as if the whole mass of the sphere
were collected at its center. * If the height of the particle above
the earth's surface is small compared with the radius of the earth,
the law of attraction becomes, since r in (69-1) is sensibly con-
stant and equal to the radius of the earth, t
(69-2) F = mg,
where g is a new constant called the acceleration due to gravity.
Its value in the c.g.s. system is approximately 980 cm. per second
per second and in the f .p.s. system 32.2 ft. per second per second.
Thus, the differential equation of the falling body can be
written as
(69-3) g = g,
where s is the distance traveled by the body and t is the time in
seconds. Integration of (69-3) gives
* In this connection, see Sees. 16 and 66.
§70 ORDINARY DIFFERENTIAL EQUATIONS 233
(69-4) ~ = gt + t>o;
and, since the velocity v is equal to ds/dt, (69-4) may be written
v = gt + v0,
where VQ is the constant of integration so chosen as to equal the
initial velocity, that is, the value of v at the time t = 0.
Integrating (69-4) gives
(69-5) s = y2gP + vQt + s0,
where SQ is the distance of the body from the point of reference
at the time t = 0. Equation (69-5) furnishes all the desired
information about the freely falling body.
70. Simple Harmonic Motion. Simple harmonic motion is the
most important form of periodic motion. It represents a
linear vibration of such a sort that the vibrating particle is
accelerated toward the center of its path in such a way
that the acceleration is proportional to the displacement of the
particle from the center. If the displacement of the particle
from its central position is denoted by #, the definition of simple
harmonic motion demands that
r/2r
(70-1) *Z = -„,*,
where a?2 is a constant of proportionality and the negative sign
signifies that the acceleration is directed oppositely to the dis-
placement x.
In order to find the equation of motion, that is, the displace-
ment of the particle in terms of the time t, multiply both sides
fir
of (70-1) by 2 ~ and obtain
The left-hand side of (70-2) is the derivative of (dx/dt)2, and
integration yields
•
(£)'--.
where the constant of integration is written for convenience in
234 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §71
the form c2, for it must be positive; otherwise, the velocity
dx/dt will be imaginary.
Extracting the square root and solving for dt give
dt =
dx
which upon integration becomes
• 1 * i
- sin"1 — = t +
or
(70-3)
x = A sin
+
where A = c/u and B = Cio>. The period of the motion,
T7 = 27T/CO, is independent of tho amplitude A.
It will be seen in the next section that (70-3) approximately
represents the behavior of a simple pendulum.
71. Simple Pendulum. Let P be a position of the bob of a
simple pendulum of mass m and of length I (Fig. 66), and let 0
be the angle, measured in radians, made by OP
with the position of equilibrium OQ. Denote the
tangential acceleration by d*s/dt2, where s repre-
sents the displacement, considered positive to the
right of OQ.
The acceleration d2s/dt* along the path of the
bob is produced by the tangential component of
the force of gravity mg, so that its magnitude is
mg sin 0. Since the velocity of the bob is decreas-
ing when the bob is moving to the right of its
position of equilibrium OQ, the acceleration will
be negative. Hence, since force is equal to the product of mass
and acceleration, one can write
FIG. 66.
(71-1)
dt*
The normal component of the force of gravity acts along OP
and is balanced by the reaction of the string (Newton's third
law of motion).
Remembering that s = 10, (71-1) can be written as
(71-2)
§71 ORDINARY DIFFERENTIAL EQUATIONS 235
and if the angle 0 is so small that* sin 0 can be replaced by 6,
This equation is precisely of the form (70-1), and its general
solution is
(71-4) 0 = Ci sin (ut + c2),
where Ci and c2 are arbitrary constants and co2 = g/l.
However, from physical considerations it is clear that there
is nothing arbitrary in the behavior of the pendulum. Moreover,
it is known that, if the pendulum bob is held initially at an angle
a and then released without receiving any impulse, the pendulum
will vibrate in a perfectly definite manner, so that it must be
possible to calculate the position of the bob at any later time t.
These remarks concerning the initial position of the pendulum
bob and the fact that the bob was released with zero velocity
can be stated mathematically as follows: If the time at which
the pendulum was released is denoted by t = 0, then
SO = a when t = 0,
^ = 0 when t = 0.
at
Therefore, the general solution (71-4) of (71-3) must satisfy the
initial conditions (71-5). Substituting the first of these initial
conditions in (71-4) gives
(71-6) a = Ci sin c2.
Differentiation of (71-4) with respect to t shows that
dO . . . N
^ = Cio> cos (co£ + c2),
and therefore the second initial condition yields
0 = Cico cos c2,
which is satisfied if c2 = ir/2. Substituting this value of c2 in
(71-6) gives Ci = a. Thus, the particular solution of (71-3) that
satisfies the initial conditions is
6 = a sin ( co£ + ^ J = a cos ut.
* See Prob. 11, Sec. 13.
236 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §71
Naturally, a different choice of the initial conditions would lead
to different values for c\ and 0%.
The solution of the problem of a simple pendulum that was
just obtained was based on the assumption that 6 was sufficiently
small to permit the replacement of sin 6 by 6. If this is not the
case, the problem is somewhat more difficult. In order to
solve (71-2), multiply both sides by 2 -^> obtaining
ddd*6 __ __?0 ^.
*Ttw ~~ i *m°dt
Integration gives
(dO\ __ 2g
\dtj ~ I
cos e + c.
i
Since dB/dt = 0 when 0 = a,
0 = -~ cos a + C
dt) I
and
The angular velocity is given by dQ/dt] and since the linear
velocity is / -T:I the velocity in the path at the lowest point is
u/t
- (cos e — cos a) I
0=0 = \/2gl(l — cos a).
It may be observed that this is the same velocity that would
have been acquired if the bob had fallen freely under the force
of gravity through the same difference in level, for v =
and h = 1(1 — cos a).
Integrating (71-7) yields
(71-8) ' /T r de
\/cos 0 — cos a
which gives the formula for determining the time required for the
bob to move from the initial position to any other.
§71 ORDINARY DIFFERENTIAL EQUATIONS 237
If the lowest position of the bob is chosen as the initial position,
then 6 = 0 when t = 0, and (71-8) becomes
(Tl-9) < - Jl •"" M
o ^/cos 0 — cos a.
where 0 ^ 0i ^ a.
In order to evaluate (71-9), first reduce the integral to a more
A
convenient form by means of the relation cos 0=1—2 sin2 •=•
£
Then,
Let
then
. 0 . a .
sin p: = sin -jr sin
jn ' «
cos 2 • Q «^ = sin s cos
and
2 sin ^ cos (p d<p 2 sin ~ cos <p d<p
cos H x/l — sin2 ^ sin2 ^
^s \ A
Substitution of these expressions in (71-10) gives
rr (*<p, 2 sin jj cos
t - — \ *
\2fli Jo
{ sin2 ^ — sin2 ^ sin2 ^ j ^/l — sin2 ^ sin2 <p
or
I-J^
V1 - sin2 1 sir
If the time involved is the time required for the completion of
one-quarter of the vibration, then 0i = a and hence y\ = ir/2.
238 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §71
The entire period is then
" *2
T =
-%/!"" sin2 ~ sin2 <p
9 Jo VI - W sin2 <?
where fc2 = sin2 -^
If (1 — k2 sin2 <p)~W is expanded by the binomial theorem, so
that
T = ±J- Pdv(l+h
\g Jo \ *
term-by-term integration* gives
It may be noted that the period is a function of the amplitude,
which was not true in the case of simple harmonic motion.
N
FIG 67.
A reference to Sec. 14 shows that the period of a simple
pendulum is expressible as an elliptic integral of the first kind.
e Note Wallis's formula
(n - l)(n - 3)
2 or 1
P sinn e dO = f2 cosn e dO = ^ . ^ rtx — ^ :: — a,
Jo Jo n(n — 2) • • • 2 or 1 '
where a. » 1 when n is odd, a = Tr/2 when n is even.
§72 ORDINARY DIFFERENTIAL EQUATIONS 239
72. Further Examples of Derivation of Differential Equations.
1. The Slipping of a Belt on a Pulley. Let T0 and TI be the
tensions of the belt (Fig. 67) at the points A and B. Consider
an element of the belt of length As, which has end points P and Q
and subtends an angle A0 at 0. Let the tension at P be T and at
Q be T + AT, and let the normal pressure per unit of length of the
arc be p, so that the total normal force on the element of arc
As is p As. If the angle A0 is assumed to be small, the normal
pressure may be thought of as acting in the direction of the
line ON, which bisects the angle A0. From the definition of the
coefficient of friction /z, it follows that the frictional force is equal
to the product of M by the normal pressure, so that the frictional
force on PQ is up As, and, since A0 is small, this frictional force
may be assumed to act at right angles to ON. If it is assumed
that the belt is at the point of slipping, the components of force
along ON must balance. Hence,
T sin ^ + (T + AT) sin ^ = p As
or
(72-1) (2T + AT) sin ~? - p-As.
Similarly, by equating the forces acting at right angles to ON,
(T + AT) cos ^ - T cos y = Mp As
or
(72-2) AT cos — = MP As.
Z
Eliminating p As between (72-1) and (72-2) leads to
,70 o\ 2T + AT . A0 1
(72-3) — ~= — tan — = —
Solving (72-3) for AT gives
*m ,. A0 2T/i
AT = tan -^r • — — >
Z 1 A0
1 — M tan -^
and dividing both members of this equation by A0 leads to
240 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §72
A0
AT
A0 A0 A A0
"2 1 ~ M tan "2
The limit of this expression as A0 — > 0 is
(72-4) f = ,r,
snce
,. tan a .
lim - = 1.
Separating the variables in (72-4) yields
?-,*.
which, upon integration, becomes
log T = fjid + c
or
(72-5) T = c^B.
The arbitrary constant Ci that enters into the solution of the
differential equation can be determined from the initial condition
T = jT0 when 6 = 0. Substituting these values in (72-5) gives
T =
so that the tension TI corresponding to the angle of lap a is
PROBLEM
Find the tensions T\ in the foregoing illustration when To = 100,
M = K> and the angles of lap are Tr/2, %TT, and TT radians.
2. Elastic Curve. Consider a horizontal beam under the
action of vertical loads. It is assumed that all the forces acting
on the beam lie in the plane containing the central axis of the
beam. Choose the #-axis along the central axis of the beam in
undeformed state and the positive y-axis down (Fig. 68). Under
the action of external forces Ft the beam will be bent and its
central axis deformed. The deformed central axis, shown in
ORDINARY DIFFERENTIAL EQUATIONS
241
the figure by the dotted line, is known as the elastic curve, and it
is an important problem in the theory of elasticity to determine
its shape.
It can be shown* that a beam made of elastic material that
obeys Hooke's law is deformed in such a
way that the curvature K of the elastic
curve is proportional to the bending
moment M. In fact,
(72-6) K =
M
w
FIG. 68.
where E is Young's modulus, / is the moment of inertia of the
cross section of the beam about a horizontal line passing through
the centroid of the section and lying in the plane of the cross
section, and y is the ordinate of the elastic curve. The important
relation (72-6) bears the name of the Bernoulh-Euler law.
The bending moment M in any cross section of the beam is
equal to the algebraic sum of the moments of all the forces F,
acting on one side of the section. The moments of the forces F,
are taken about a horizontal line lying in the cross section in
question.
If the deflection of the beam is small, the slope of the elastic
curve is also small, so that one may neglect the square of dy/dx
in the formula for curvature. Thus, for
small deflections the formula (72-6) can
• be written as
(72"7) ~dtf = ™
M_
El
FIG. 69.
As an illustration of the application of
this formula, consider a cantilever beam
of length I, which is built in at the left end and which carries a
load W on its free end (Fig. 69). The weight of the beam is
assumed negligible in comparison with the magnitude of the load
TF, so that the moment M in any cross section at a distance x
from the built-in end is
M = W(l - x).
* See TIMOSHENKO, S., Theory of Elasticity, p. 41; LOVE, A. E. H., A
Treatise on the Mathematical Theory of Elasticity, 4th ed., pp. 129-130.
242 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §72
When this expression is substituted in (72-7), there results
dtf = m (l " **'
and integrating gives
W fix* x*
The constants of integration, ci and c2, are easily evaluated
from the boundary conditions
y = 0, when x = 0,
~~ = 0, when re = 0,
the first of which expresses the fact that the displacement at
the built-in end is zero and the second that the slope of the
elastic curve is zero when x = 0. It is easily checked that these
boundary conditions require that
W
y " 2EI
so that the displacement d at the free end is
Wl3
3EI
PROBLEM
A beam of length / is freely supported at its ends and is loaded in the
middle by a concentrated vertical load W,
which is large in comparison with the weight
jv^ | ^ of the beam. Show that the maximum
deflection is one-sixteenth of that of the
cantilever beam discussed above. Hint:
From symmetry, it is clear that the behavior
70t "~ of this beam is the same as that of the
cantilever beam of length 1/2 which is
loaded by a concentrated load of magnitude W/2 at its free end (Fig.
70).
3. Cable Supporting a Horizontal Roadway. Let a cable that
supports a horizontal roadway be suspended from two points
A and B (Fig. 71). It will be assumed that the load on the
roadway is so large compared with the weight of the cable that
§72
ORDINARY DIFFERENTIAL EQUATIONS
243
T+AT
the weight of the cable can be neglected. The problem is to
determine the shape assumed by the cable.
Denote the tension at the point P of the cable by T and that
at the point Q by T + AT, and let w be the load per foot run.
Since the cable is in equilibrium, the horizontal and vertical
components of the forces acting on any portion As of the cable
must balance. Thus, equat-
ing the horizontal and ver- y
tical components gives a
system of two equations
(72-8) T cos 0
= (T + AT) cos (0 + A0)
and
(72-9) T sin0 = —wAx
+ (T + AT) sin (0 + A0).
Dividing (72-9) by (72-8) ~~
gives
(72-10) tan 0 = tan (0 + A0)
FIG. 71.
W Ax
But (72-8) does not depend on the magnitude of As and, since
As is arbitrary in size, it appears that the horizontal component
of the tension at any point of the cable is a constant, say TO.
Substituting this value in the right-hand member of (72-10)
and rearranging give
1J) AT*
tan (6 + A0) - tan 0 = =^=f
cr
(72-11)
tan (0 + A0) — tan 0 = w_ Az
A0 ~~ To A0'
Tho left-hand member of (72-11) is the difference quotient, and
its limit as A0 is made to approach zero is the derivative of
tan 0. Hence, passing to the limit gives
,*
w_ dx
''YQTe
(72-12)
2
SGC
Recalling that tan 0 = —•> so that 0 = tan"1 •—; it follows that
ax ax
dx 1 + (dy/dxY
244 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §72
Moreover,
sec2 0=1+1^
\dx
Substituting from these two expressions in (72-12) leads to the
differential equation of the curve assumed by the cable, namely,
V - w.
d& ~ ri
If (72-13) is integrated twice with respect to #, one obtains the
desired equation of the curve,
ID
(72-14) y = _ 32 + ClX + C2;
which is the equation of a parabola. The arbitrary constants Ci
and C2 can be determined by substituting in (72-14) the coordi-
nates of the points A and B.
If the lowest point of the cable is chosen as the origin of the
coordinate system, the equation of the parabola becomes
<ii)
(72-15) y = WoX*-
The length of any portion of the cable can easily be calculated
with the aid of (72-15).
PROBLEMS
1. Find the length of the parabolic cable when the latter supports a
roadway which is I ft. long. Express the length of the cable in an
infinite series in powers of L Hint: Expand the integrand in the
expression for the length of the cable.
2. Find an approximate expression for the sag d in terms of the
length I by using the first two terms of the infinite series expansion that
was obtained in Prob. 1.
4. Uniform Flexible Cable Hanging under Its Own Weight. Let
a flexible cable (Fig. 72) be suspended from two points A and B.
Denote the weight per \Hiit length of the cable by w, and con-
sider the forces acting on the element of cable As. As in the
preceding example, the horizontal and vertical components of
force must balance, for the cable is in equilibrium. If the tension
at P is denoted by T and that at Q by T + AT7, it follows that
T cos 6 = (T + AT) cos (0 + A0)
§72
and
ORDINARY DIFFERENTIAL EQUATIONS
245
T sin B = (T + AT) sin (6 + A0) - w As.
Dividing the second of these equations by the first gives
w As
tan 6 = tan (0 + A0) -
(T + AT) cos (0 + A0)
This equation has the structure of Eq. (72-10), and an analysis
in every respect similar to that outlined in the preceding illus-
T+AT
FIG. 72.
tration leads to the equation
(72-16)
w ds
sec° B = f0 35'
where To is the tension at the lowest point of the cable. Since
ds ds/dx
dO dd/dx
where
dx
^ and
1 +
and since sec2 0=1 + (dy/dx)2, it follows upon substitution
in (72-16) that the differential equation of the curve assumed by
the cable is
(72-17)
dx*
246 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §72
If dy/dx is replaced by w, (72-17) becomes
du __ w /T~^T — 2
Tx ~ n V1 + u>
or
rfu ^ 7
Integrating this equation gives
log (u H
This differential equation can be solved by the following
device: Taking the reciprocal of (72-18), one obtains
and rationalizing the denominator gives
When (72-19) is subtracted from (72-18), there results
rfy if £«+«, -(f *+«»
^ . -^r. _ e Vr.
and integration gives
The constants c\ and eg can be determined from the condition
that the curve passes through the points A and B, whose coordi-
nates are assumed to be known.
If the constants c\ and cz are chosen to be equal to zero, then
the lowest point of the curve is at (0, TQ/W), and the equation
§73 ORDINARY DIFFERENTIAL EQUATIONS 247
of the curve assumed by the cable has the form
(72-20) y =
A curve whose equation has the form (72-20) is called a catenary.
PROBLEM
Find the length of the catenary between the limits 0 and v.
73. Hyperbolic Functions. Combinations of exponential func-
tions analogous to the one that appears in (72-20) are of such
frequent occurrence in applied mathematics that it has been
found convenient to give them a special
name. The function %(ex + e~x) is
called the hyperbolic cosine of x and is
denoted by
(73-1) cosh x - y^e* + <r*).
The derivative of cosh x is equal to
%(ex — e~x) and is called the hyper-
bolic sine of x. Thus,
(73-2) sinh x - %(cx - e~x)
These functions are called hyperbolic because they boar rela-
tions to the rectangular hyperbola x2 — y2 = a2 that are very
similar to those borne by the circular functions to the circle
x2 -f i/2 = a2. Thus, consider
the equation of a circle (Fig. 73)
whose parametric equations are
x = a cos t
and
y = a sin t.
The equation of a rectangular
FIG. 74. hyperbola (Fig. 74) is
(73-3) x2 - t/2 = a2,
and the reader can readily show with the aid of the definitions
248 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §73
(73-1) and (73-2) that (73-3) can be written in a parametric form
as
x = a cosh tj
= a sinh t.
(73-4)
It will be shown next that the parameter t can be interpreted
for the circle and the hyperbola in a similar way.
The area u of the circular sector OAP (Fig. 73) is
u
2u
so that
On the other hand, the area of the hyperbolic sector OAP (Fig.
74) is given by
(73-5) u =
~~ / *N/#2 ~ #2 dxj
where the first term in (73-5) represents the area of the triangle
OBP.
Integrating (73-5) gives
x + -\/x2 — a2 Q? 1 x + y
= 9" log
2 ™* ~^T)
.
log
x + y 2u
- =
so that
and
(73-6)
Also, since a;2 — y2 = a2, it follows that
r — 11 -—
(73-7) £--? = e a'.
Adding and subtracting (73-6) and (73-7) lead to
(73-8)
x = a
.2u
= a cosh -r;
a2
2?*
2u
= a
= a sinh
2u
which are precisely Eqs. (73-4) with t = 2u/o2.
§73 ORDINARY DIFFERENTIAL EQUATIONS 249
From (73-8), it is clear that '
x . 2u . y . . 2u
- = cosh -r- and - = sinh — r
a a2 a a2
and a reference to Fig. 73 shows that
x 2u , y . 2u
- = cos —r and - = sin -=-
a a2 a a2
Therefore, the circular functions may be defined by means of
certain ratios involving the coordinates of the point P(x, y) on
the circle x2 + y2 = a2, whereas the hyperbolic functions are
expressed as ratios involving the coordinates of the point P(x, y)
on the hyperbola x2 — y2 — a2.
The definitions of the hyperbolic tangent, hyperbolic cotangent,
hyperbolic secant, and hyperbolic cosecant are as follows :
, sinh x
tanh x = — ; — >
cosh x
COth X — ; ;
tanh x
sech x = — ; — )
cosh x
csch x =
sinh x
The inverse hyperbolic functions are defined in a way similar
to that used in defining the inverse circular functions. Thus, if
y = tanh x,
then
x = tanh"1 y,
which is read " the inverse hyperbolic tangent of y." The definition
of the remaining inverse hyperbolic functions is similar. There
are some interesting relations that connect these inverse hyper-
bolic functions with the logarithmic functions.*
It will be recalled that the expansion in Maclaurin's series for eu
is
(73-9) «..! + „ + «; + «;+...,
* See Probs. 5 and 7 at the end of this section.
250 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §73
so that
(73-10) *• = 1 + x + ~ + ~ + • • •
and
(73-11) e- = 1 - x + £ - £ + • • • .
Subtracting (73-11) from (73-10) gives
so that
pX p — X /y»3 /y«5
(73-12) sinh x m —$— = * + |-, + |-, + • • • .
On the other hand, addition of (73-10) and (73-11) shows that
/?-£ I p — X /v»2 /y»4
(73-13) cosh x = e—±— = 1 + |l + |l + ' ' ' •
Moreover, if it is assumed that (73-9) holds for complex num-
bers as well as for real numbers, then
(Yr")2
(73-14) e» = 1 + ix + ^jf-
and
(73-15) ^ = l-ix +
where i s= \/— 1. Adding (73-14) and (73-15) and simplifying
show that
T2 T4 T6
\
),
which is recognized to be the series for cos x multiplied by 2.
Thus,
pix JL. p—ix
(73-16) cos x = — —
It is readily verified that subtraction of (73-15) from (73-14)
leads to the formula
fix _. p-ix
(73-17) sin x = - --
§73 ORDINARY DIFFERENTIAL EQUATIONS 251
By combining (73-16) with (73-17) there result two interesting
relations,
cos x + i sin x = elx and cos x — i sin x = e
= ~lx
which are frequently used in various investigations in applied
mathematics. These relations are known as the Euler formulas.
The following table exhibits the formal analogy hat exists
between the circular and hyperbolic functions. The relations
that are given for hyperbolic functions can be established readily
from the definitions for the hyperbolic sine and the hyperbolic
cosine.
Circular Functions Hyperbolic Functions
sin x = 2~ (etx — e~tar) sinh x = ^ (ex — e~x)
cos z = 2 (e'x + «"**) cosh x = ^ (ex + e~x)
elx — e~*x , , ex — «~*
tan x = —t - : - -v tanh a; = — — - -
^(elx + e~%x) ex + e~*
cot # = r -- coth x =
• -- V/vsuii A* — 7 i
tan x tanh a;
/v.2 /v«4 /r2 /v«4
cos x = 1 - 2j + j| - cosh x = 1 + 2j + jj + • • •
sin2 x + cos2 x = 1 cosh2 x — smh2 x = 1
1 4- tan2 x — sec2 x I — tanh2 x = sech2 x
sin 2z = 2 sin z cos x smh 2x = 2 smh # cosh x
cos 2x = cos2 x — sin2 # cosh 2x = cosh2 x -f sinh2 a;
sin (x ± y) — sin x cos i/ sinh (x ± ?/) = sinh x cosh t/ ± cosh x smh ?/
± cos x sin i/
</ sin x d sinh x ,
cos x — T - = cosh x
, — \J\JO As i
dx dx
d cos x _ _ . d cosh x
dx ~ dx
= smh a;
d tan x 0 d tanh x , ,
— 3 - = sec2 x - -5 - = sech2 x
dx dx
Example 1. A telephone wire (Fig. 75) weighing 8 Ib. per 100 ft. is
stretched between two poles 200 ft. apart. If the sag is 1 ft., find tlu»
tension in the wire.
Note that
)-
where a = T0/w.
252 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §73
The vertical component of the tension is clearly equal to ws, where
« is the length of the wire. But the length of the catenary between the
points whose abscissas are 0 and x is
s —
where y = a cosh — Substituting for
Y"01*4* dy/dx gives
•
-f:
FIG. 75.
so that the vertical component of tension is
1 + sinh2 - dx
u j - i.
co ~ ~ a 81 "'
Ty — ws = wa sinh ->
and the total tension at any point is
T = V2V +
= wa J
l + sinh2
wa cosh -
a
wy.
At the point of support, y = a + d, so that T = w(a + d). Since d is
usually small, the tension in the wire is nearly constant and approxi-
mately equal to TV
If the wire is very taut and the distance between the poles is not large,
r-2*
- a - 25'
When x = Z/2, where Z is the distance between the poles, and d is the
sag> y — a = d and
so that
* The symbol a = 6 is used to signify that a is approximately equal to 6.
§73 ORDINARY DIFFERENTIAL EQUATIONS 253
or
T° = 8d'
Substituting the numerical values for w, I, and d gives for the value
of the tension at the lowest point
- (O.Q8)(200)2 _
To== (8)'(1) -4001b'
Example 2. A parachute, supporting a mass m, is falling from a
distance h above the ground. Determine the velocity with which it
strikes the ground if the air resistance is proportional to the square of
the velocity.
If the air resistance be denoted by R, then
R = kv2,
where fc is a proportionality constant depending upon the design of
the parachute. The force acting downward is
d2s dv
which is equal to mg — kv2 Hence,
dv
m -r. = mg — kv*
or
g-r = g(\ - oV),
where a2 = k/gm. Integrating
f dv r
J i -.aV = 9J *
gives
1 i 1 + w
If v = 0 when t = 0, it follows that c\ = 0. The integrated expres-
sion then simplifies to
l + av
— av
254 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §73
or
It is easily shown that
lim tanh t — 1,
and it follows that the terminal velocity is vt = \/gm/k.
But ds/dt = v, so that
1 f
s = - I tanh agt dt
a J
= —£- log cosh agl + c2;
and since s = 0 when t = 0, c2 = 0. Hence,
m i i W *
s = 7T log cosh A/ — £.
PROBLEMS
1. A wire is stretched between two supports 100 ft. apart. If
the weight of the wire is 10 Ib. per 100 ft. and the tension in the wire is
300 Ib., find the amount of sag at the middle.
2. Newton's law of cooling states that the rate of decrease of the
difference in temperature of a body surrounded by a medium of con-
stant temperature is proportional to the difference between the tempera-
ture of the body and that of the medium, that is,
dO _
dt
Find the temperature 0 of the body at any time t, if the initial tempera-
ture is 0i.
3. If a wire weighing w Ib. per unit of length is stretched between two
supports I units apart, show that the length of the wire is approximately
where T is the tension.
4. Show that any complex number a + U can be put in the form
a + bi = re6*, where r = \/a2 + &2 and 0 = tan"1 -•
§73
ORDINARY DIFFERENTIAL EQUATIONS
255
5. If y = sinh x, then x is called the inverse hyperbolic sine and is de-
noted by x = sinh-1 y. Prove that x = sinh-1 y = log (y + vV + 1).
6. Establish the formulas for hyperbolic functions given in the table
of Sec. 73.
7. Establish the following formulas:
(a) d sinh u = cosh u du;
(b) d cosh u = sinh u du]
(c) d tanh u = sech2 u du;
(d) d coth u — — csch2 u du;
(e) d sech u = — sech u tanh u du\
(/) d csch i« = — csch u coth
(<;) d sinh"1 u =
(fi) d cosh"1 u —
(i) d tanh-1 u =
(j) d coth-1 u =
du
du
du
1 - u2'
du
u \/\ - u2'
du
(k) d sech-1 u = —
(I) d csch"1 u = —
u VI +
(w) cosh"1 y — log (y + V?/2 — 1) = sinh"1 •
(n) sinh-1 y = log (y + Vl/2 H~ 1) = cosh"1 •
(o) tanh-1 y = g log 1 _ if 2/2 < 1;
(p) coth-1 y = 5 log
(#) seen"1 y
(r) csch-1 y
log
+ 1;
< 1;
8. Plot the graphs of the hyperbolic functions.
9. A man and a parachute, weighing w lb., fall from rest under the
force of gravity. If the resistance of the air is assumed to be propor-
tional to the speed v and if the limiting speed is v0, find the expression for
the speed as a function of the time t.
Hint:
w dv
256 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §74
10. A man and a parachute are falling with the speed of 100 ft. per
second at the instant the parachute is opened. What is the speed of
the man at the end of t sec. if the air resistance is proportional to the
square of the speed?
11. It can be established that the steady flow of heat across a
large wall is proportional to the space rate of decrease of temperature
0 across the wall and to the area A of the wall, that is,
where x is the distance from one of the faces of the wall and Q is the
constant quantity of heat passing through the wall. The constant k
(thermal conductivity) depends on the properties of the material.
Integrate this equation and calculate the amount of heat per square
centimeter passing through a refrigerator wall, if the thickness of the
wall is 6 cm. and the temperature inside the refrigerator is 0°C., while
outside it is 2(J°C. Assume k = 0.0002.
12. A tank contains initially v gaL of brine holding XQ lb. of salt in
solution. A salt solution containing w lb. of salt per gallon enters the
tank at the rate of r gal. per minute; and the mixture, which is kept
uniform by stirring, leaves the tank at the same rate. What is the
concentration of the brine at the end of t min. ?
Hint: Let x denote the amount of salt present at the end of t min.;
then, at a later instant t + A£, the change in the quantity of salt is
A# = wr At — (x/v)r At. Hence, dx/dt — wr — xr/v = (r/v)(wv — x).
74. First-order Differential Equations. Generally speaking,
the problem of solving differential equations is a very difficult
one. There are very few types of equations whose solutions can
be written down at once; in practice, special methods of solu-
tion, suitable to the particular problem under consideration,
have to be depended upon. Seeking special methods of solution
is a difficult task, and the mathematician, at present at least, is
almost entirely restricted to a consideration of linear differential
equations. Very little is known concerning the solution of non-
linear differential equations. Even such a simple-appearing
first-order equation as
cannot be solved in general; that is, there are no formulas avail-
able for solving a non-linear differential equation of the first order.
However," it is possible to classify some of the first-order non-
§76 ORDINARY DIFFERENTIAL EQUATIONS 257
linear differential equations according to several types and to
indicate the special methods of solution suitable for each of these
types. The next ten sections will be concerned with the solutions
of the special types of non-linear differential equations that are of
common occurrence in practice. The remainder of the chapter
will be devoted to the general methods of solution of the impor-
tant types of linear differential equations.
75. Equations with Separable Variables. If the given differ-
ential equation
can be put in the form
/i Or) dx + /2(y) dy = 0,
where /i(o;) is a function of x only and fa(y) is a function of y
only, the equation is said to be an equation with separable
variables. Such an equation is easily integrable, and its general
solution is
ffi(x) dx + f My) dy = c,
where c is an arbitrary constant. In order to obtain an explicit
solution, all that is necessary is to perform the indicated
integrations.
Example. Solve
dy , .
~T~ ~r &xy = &xy •
This can be written as
~dx "*" e*^y "" y ' ^
or
Integration gives
log . __ ^ + e* = c,
which is the general solution required.
258 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §75
PROBLEMS
Solve the following differential equations:
1. VI - & dy + V 1 - yz dx = 0.
dy _ sin2 x
dx ~~ sin y
4. sin x cos2 y dx + cos2 x dy = 0.
6. Vl + x dy - (1 + y2) dx = 0
0.
6. e* VI ~ y2 dx + - dy
7.
dy I + y
dx 1 + x
8. e« ^ + y - y2 = 0.
9. sinh x dy + cosh y dx — 0.
dx
dx
12.
13.
og 2/ :/ + tan z sec2 x.
4?/2) dx + 3yx* dy = 0.
sin
+ (1 - e*) dy = 0.
xy
-
* dx ~ x(y - 1)'
16. (1 + x2) dy - (1 + y2) dx = 0.
17 ^ - y8 + 2y + i
•"' dx "" x2 - 2x + T
18. x2(l + y)dy + y*(x - 1) dx = 0.
19. y(l -y)dx- (x+ 1) rfy = 0.
dy _ x(l + y2)
^- dx " y(l + x2)'
21. (y2 - xy) dx + x2 dy = 0.
22. Let A be the amount of a substance at the beginning of a chemical
reaction, and let x be the amount of the substance entered in the reaction
after t sec. Then, the simple law of chemical reaction states that the
rate of change of the substance is proportional to the amount of the sub-
stance remaining; that is, dx/dt = c(A — x), where c is a constant
depending on the reaction. Show that x = A(l — e~cO.
§76 ORDINARY DIFFERENTIAL EQUATIONS 259
23. Let a solution contain two substances whose amounts expressed
in gram molecules, at the beginning of a reaction, are A and B. If an
equal amount x of both substances has changed at the time t, then the
amounts of the substances remaining are A — x and B — x. The basic
law of chemical reactions states that the rate of change is proportional
to the amounts of the substances remaining; that is,
ft = k(A - x)(B - x).
Solve this equation under the hypothesis that x = 0 when t = 0.
Discuss the case when A — B.
76. Homogeneous Differential Equations. It will be recalled*
that a function f(x, y), of the two variables x and y, is said to be
homogeneous of degree n provided that
f(\x, \y) ^ \»f(x, y).
Thus,
f(x, y) = z* + x*y + y3
is a homogeneous function of degree 3, and
f(x, y) = x2 sin ^ + xy
y
is a homogeneous function of degree 2.
If the differential equation is of the form
(76-1) /!(», y) dx + fr(x, y) dy = 0,
where /i (x, y) and/2(x, y) are homogeneous functions of the same
degree, then (76-1) can be written in the form
where <p(x, y) is a homogeneous function of degree zero, that is,
<p(\x, \y) 35 \° <p(xy y) = <p(x, y).
If X is set equal to l/x, then
<p(x, y) =
which shows that a homogeneous function of degree zero can
always be expressed as a function of y/x. This suggests making-
* See Sec. 40.
260 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §76
the substitution y/x = v. Then, since y = vx,
dy dv ,
-/- = 3- a; + v.
dx ax
Substituting this value of dy/dx in (76-2) gives
x
This equation is of the type considered in Sec. 75. Separating
the variables leads to
dv dx
<p(l,v) - v x
which can be integrated at once to give
F(v, x, c) = 0.
Since v = y/x, the general solution of (76-1) is
F(x'X
An equation of the form
dy _ a\x + azy +
dx bix + b^y + 63
can be reduced to the solution of a homogeneous equation by a
change of variable. This is indicated in detail in Prob. 11 at
the end of this section.
Example. Solve
o , 9dy dy
v. + a.,_ = xys. .
This equation can be put in the form
y*dx + (x* - xy) dy = 0,
which is of the type (76-1). By setting y = vx and dy = v dx + x dv,
the equation becomes
(vxY dx + (x* - vx*)(v dx + x dv) = 0.
This reduces to
v dx + x(l — v) dv = 0
§76 ORDINARY DIFFERENTIAL EQUATIONS 261
and, upon separation of the variables, to
dx . 1 — v . A
dv = o.
x v
Integration yields
log x + log v — v = c
or
which simplifies to
log y - jj- = c.
PROBLEMS
Solve the following differential equations:
1. (x2 + y2) dy + 2xy dx = 0.
2' xdx~y " V^17?.
4. (x + T/) = x - y.
5. x2y dx - (x3 - ?/) dy = 0.
x —
8. x(\fxy + y) dx - x2 dy = 0.
dy __ y* - x Vx* - y*_
j •— •
11. Discuss the problem of transforming the differential equation
dy _ a\x + a2y + a?
into a homogeneous equation by the change of variable x = x' + h and
y = ?/ + fc. Determine the values of h and fc for which the original
262 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §77
equation is transformed into
<ty _ a\xf + a2yr
dx' b&' + b*yf
and solve this equation. If aj)z — ajb\ = 0, set a\x + a^y = z.
12. (x* - xy) dy + y* dx = 0.
13. (y2 - z2) ch/ + 2zi/ dx = 0.
dy 1 + 2s + y
140 dx 1 - 2x - y
16. (a? - y + 1) efo + (x + y - 1) dy = 0.
16. i/2 dx + (xy + x*) dy = 0.
17. (2x3y - ?/) dx + (Zxif - z4) dy = 0.
18. (z2 + 2/2) rfx + Bxy dy = 0.
19. (x2 + y2) dx - xy dy = 0.
20. (x + y) dy - (x - y) dx = 0.
77. Exact Differential Equations. It was shown in Sec. 63
that the necessary and sufficient condition that the expression
P(x, y} dx '+ Q(x, y) dy
be an exact differential of some function F(x, y) is that
m \\ dP dQ
(77-1} ^ = ^'
where these partial derivatives are continuous functions.
Consider now the differential equation
(77-2) P(x, y) dx + Q(x, y) dy = 0,
and suppose that the functions P(x, y) and Q(x, y) satisfy the
condition (77-1), so that there exists a function F(x, y) such that
,v dF , , dF ,
dF = tedx + djdy
= P(x, y} dx + Q(x, y) dy.
Such a differential equation is called an exact differential equation.
It is clear that the function
where c is an arbitrary constant, will be a solution of (77-2).
An explicit form of the function F(x, y) will be obtained next.
§77 ORDINARY DIFFERENTIAL EQUATIONS 263
By hypothesis the condition (77-1) is satisfied so that one can
write
(77-3) — = P(x, y) and — = Q(x, y).
ox oy
Now, the first of these equations will surely be satisfied by the
expression
(77-4) F(x, y) = / P(x, y) dx + /(</),
where the y appearing under the integral sign is treated as a
parameter and f(y) is an arbitrary function of y alone. The
function f(y) will be determined next, in such a way that (77-4)
satisfies the second of Eqs. (77-3).
Differentiating (77-4) with respect to y and equating -the
result to Q(x, y) give
<W
dy
so that
(77-5) j| = Q(x, y) - ± J P(x, y) dx.
Hence,
(77-6) f(y) = J [<2(x, tf) - ^ J P(«, J/) dx ] dy.
Substitution of (77-6) in (77-4) gives the explicit formula
(77-7) F(x, y) = J P(x, y) dx + J [<3(z, y)
y) dx \ dy.
- ± J P(x,
To illustrate the use of this formula, consider
(2xy + 1) dx + (x* + 4y) dy = 0.
Here,
¥• - ^ - 2x
dy ~ dx ~ ZX>
so that the formula (77-7) is applicable. The reader will verify
that the substitution of the expressions for P and Q in (77-7)
gives
F(x, y) = x2y + x + 2y* + c.
264 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §77
Hence, the solution is
xzy + x + 2y2 = c.
Instead of using the formula (77-7), one frequently proceeds
as follows: Since dP/dy = dQ/dx, the existence of a function
F(x, y) such that
— = 2xy + 1 and — = x2 + ty
is assured. Now, if
is integrated with respect to x} y being treated as a constant, there
results
F(x, y) = x*y + x + ci(y),
where c\(y) is not a function of x but may be a function of y,
since y was treated as a constant. Similarly, the second
condition
dF
f -*' + *'
upon integration with respect to y, gives
F(x, y) = z2s/ + 2y2 + ca(x).
Comparison of the two expressions for F(#, #) shows that if
F(x, y) = x22/ + re + 2y*,
then
— = 2xy + 1 and — = x2 +
Thus, the general solution of the given equation is
x*y + x + 2y* = c.
PROBLEMS
Integrate the following equations if they are exact:
1. (y cos xy + 2x) dx + x cos xy dy = 0.
2. (y2 + 2z?/ + 1) dx + (2xy + x2) dy = 0.
3. (e* + 1) dx + dy = 0.
4. (3x*y - y*) dx + (x9 - 3y*x) dy = 0.
§78 ORDINARY DIFFERENTIAL EQUATIONS 265
6. (3x*y - y3) dx - (z3 + 3y2z) dy * 0.
& —9 cos - dx cos - dy = 0.
a;2 a: # x y
x1
7. 2z log y dx + — dy = 0.
1 — <jy2 "V/l #2
, do; + y . dy = 0.
- x2 VI- 2/2
9. (2z + e* log y) da; H dy = 0.
10. 2x sin y dx — x'2 cos y dy = Q.
+ « ( 1 , \ , 1
12. ( 2a; + - <?x/y ) ^j ; <
V y / y2
13. sin 2y dx + 2x cos 2?y dy = 0.
14. x2(y + 1) dx - y*(x - 1) dy = 0.
15. y(l + a:2)"1 do; — tan"1 x dy === 0.
78. Integrating Factors. It is not difficult to see that every
differential equation of the type
(78-1) M(x, y) dx + N(x, y) dy = 0,
which has a solution F(x, y) = c, can be made exact by multi-
plying both members by a suitable function of x and y. For
since F(x, y) = c is a solution of (78-1),
and it follows from a comparison of (78-1) and (78-2) that
Therefore,
-^T = M(S, y)M and — =
x + Ndy) =0
is an exact equation. The function M(X, y) is termed an
grating factor. Moreover, it is clear that there is an unlimited
number of such functions for each equation. Despite this fact,
it must not be concluded that an integrating factor can always
be found easily. In simpler cases, however, the integrating
factor can be found by inspection.
Thus, in order to solve
x dy — y dx = 0,
266 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §78
which is not exact as it stands, multiply both sides by l/xy.
Then the equation becomes
dJL _ *5 = o,
y x
which is exact. Another integrating factor for this same equation
is I/a;2. Similarly, multiplication by l/y2 makes the equation
exact.
In Prob. 1 at the end of this section will be found a few of
the integrable combinations that frequently occur in practice.
Example. The differential equation
(y2 - x2) dy + 2xy dx = 0
is not an exact equation, but on rearrangement it becomes
y2 dy + 2xy dx — x2 dy = 0,
which can be made exact with the aid of the integrating factor l/y2.
The resulting equation is
2xy dx — x2 dy _
V y2 ~ »
which integrates to
+ - -
y+y~c'
, PROBLEMS
1. Verify the following:
x dy - y dx
(6) <*(log|) =
(e) Hd(x* + y2) = x dx + y dy;
(/) d(xy) = x dy + y dx.
2. Solve the following equations by finding a suitable integrating
factor:
(a) x dy — y dx + x2 dx = 0;
(6) On/2 + y) dx + (x - x2y) dy = 0;
§79 ORDINARY DIFFERENTIAL EQUATIONS 267
(c) xdy + 3ydx = xy dy,
(d) (s2 + 2/2 + 2s) <ty -2ydx = 0;
(e) xdy — y dx = xy dy,
(/) (*2 - 2/2) <fy - 2xy dx = 0;
(0) x dy — (y + log x) dx = 0.
79. Equations of the First Order in Which One of the Vari-
ables Does Not Occur Explicitly. Suppose that the dependent
variable y does not occur explicitly in the equation. The form
of the equation is then
'(£*)-*
If this equation is solved for dy/dx to obtain
^ = f(x)
dx J(x)>
then y is obtained by a simple quadrature as
y = ff(*)dx + c.
Example. Consider
Solving for dy/dx gives
and
Hence, the solutions are
2 + A/3
y -- g - * "" c = °
and
2 - \/3 o
2/ --- 2^— x2 - c = 0.
These solutions can be combined into one equation by multiplying one
by the other to give
(y - c)2 - 2x*(y - c) + Y^ = 0.
If the independent variable is missing, the equation is of the
form
F(=*>
268 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §79
Solving for dy/dx gives
or
dx 1
dy /(</)'
Integration of this equation yields
,= CtLji
Occasionally, the differential equation can be solved easily by
factoring. For example, consider
"" ' "' dx ' •
This equation can be written in the form
so that one is led to the solution of the differential equations
^-2/2 = 0 and 2^-* = 0.
dx y dx
It follows that the general solution of the given equation can be
written as
PROBLEMS
Solve the following differential equations:
*•(!)'+*-'•
' dx 1 + y* dx
§80 ORDINARY DIFFERENTIAL EQUATIONS 269
«• ' -«•+»- »•
80. Differential Equations of the Second Order. Occasionally,
it is possible to solve a differential equation of the second order by
reducing the problem to that of solving first-order equations.
Thus, if the given equation is of the form
dy\
' Tx) = °'
the substitution of p = dy/dx reduces it to
which is an equation of the first order of the type treated in
Sec. 79. If this equation is solved for p to give
P = M c),
the solution for y can be obtained at once, since p = dy/dx.
No general rules can be given for solving non-linear differential
equations, and the task must be left to the skill and ingenuity of
the student. An example of the solution of a non-linear differen-
tial equation by means of an artifice was given in Sec. 72 in
dealing with the equation of a flexible cable. Another example
may prove interesting and useful.
Example. Consider the equation
If dy/dx is replaced by p, the resulting equation is
270 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §80
Since
dp dp
*j~ — P jf~>
ax r ay
the equation can be written as
^2-2 2 + 2 = 0
or
dp 2p2 — y1
dy ~~ py
which is a homogeneous equation. Setting p = vy gives
dv 2yW - y2 2v2 - 1
which reduces to
Therefore,
and
But v = p/y, so that
and
dy _ v dv
y
* - 1
log y = log (v2 - 1)H + log
Since p = dy/dx, the last equation becomes
or
Therefore,
Combining these two solutions by multiplication gives the solution
(* + C2)'-(logC- + ^+^V = 0,
^ y /
§80 ORDINARY DIFFERENTIAL EQUATIONS 271
which can be written, also, as
/ y \2
(x + c2)2 - ( csch-1- ) = 0.
It is seen from this example that if the given differential equar
tion is of the form
(80-1) F(y, </', • - • , y<»>) = 0,
then one can introduce the new variable
P = V'
and calculate the successive derivatives as follows:
dp dp
= — L_ — _ *_
d
= — L_ — _ *_ /r)
dx dy p>
The substitution of these derivatives in (80-1) leads to a
differential equation of order n — 1. It may be possible to solve
this differential equation and obtain the general solution in
the form
p = F(y, ci, • • • , cn_i),
so that
(80-2) g = F(y, clf • • • , c,.!).
Equation (80-2) is one with separable variables.
PROBLEMS
Solve the following differential equations:
d*y
1. -7-^ + y = 0. Solve by substituting dy/dx — p, and also by
using the integrating factor 2 dy/dx.
272 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §81
5. x'g +(*-!) =0.
«•§-(!)'-'-»•
81. Gamma Functions. Consider a particle of mass m that
is moving in a straight line under the influence of an attractive
force whose intensity varies inversely as the distance of the
particle from the center of attraction. The equation of such a
motion is obtainable immediately from the definition of force
(Newton's second law). Denoting the distance from the center
of attraction by y, it follows that
d*y k
m 777? — --
dt2 y
or
where a = k/m.
This is a non-linear equation of the type
g - «»>.
which can always be solved by multiplying both sides of the
equation by 2 dy/dt and integrating. Thus,
_
dt dt* dt y
and integrating with respect to t gives
ft) = -2alog</ + c.
If the velocity of the particle is zero when y = y^ then c
2a log 2/0 and
— = — ^ I2(i loff — •
w£ ^/ y
The negative sign was chosen for the square root because y is a
§81 ORDINARY DIFFERENTIAL EQUATIONS 273
decreasing function of t. Solving for dt and integrating yield
*" dy
The integral can be put in a simpler form by making the
obvious transformation log (y$/y) = x, or y — yoe~x. If T is the
time required to reach the center of attraction, y = 0, the integral
becomes
This integral cannot be evaluated in terms of a finite number of
the elementary functions. In fact, an integral of this type led
Euler to the discovery of the so-called Gamma functions.
The remainder of this section will be concerned with the
study of the improper integral
(81-2) r(a) = f °° x?-ler*dx9 where a > 0,
which is the generalization of (81-1). It will 'be shown that
(81-2) defines an interesting function, called the Gamma function,
which provides a generalization of the factorial and which will
prove useful in the study of Bessel functions.
It is not difficult to prove* that (81-2) converges for all positive
values of a and diverges whenever a < 0. However, it is
possible to define the function T(a) for negative values of a with
the aid of the recursion formula which will be developed next.
If a > 0, then it follows from (81-2) that
(81-3) r(« + 1) = f " x«e~*dx.
•/o
Integrating the right-hand member of (81-3) by parts gives
I x«e-x dx = — xae~x °° + a f °° xa~le~-x dx
Jo o Jo
/» 00
= a I xa~le~x dx
Jo
= «r(a).
Thus,
(81-4) r(a + 1) = or(a).
*See SOKOLNIKOFF, I. S., Advanced Calculus, p. 373.
274 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §81
But
so that when a = 1 the formula (81-4) becomes
T(2) = 1 • T(l) = 1.
Setting a = 2, 3, • • • , n gives
T(3) = 2F(2) = 1-2,
T(4) = 3F(3) = 1-2-3,
T(n) = (n- l)r(n- 1) = (n - 1)!,
T(n + 1) = nT(n) = n\.
Hence, the formula (81-4) enables one to compute the values of
F(of) for all positive integral values of the argument a.
If by some means (for example, by using infinite series) the
values of F(a) are obtained for all values of a between 1 and 2,
then, with the aid of the recursion formula (81-4), the values of
T(a) are readily obtained when a lies between 2 and 3. These
values being known, it is easy to obtain T(a) where 3 < a < 4,
etc. The values of T(a) for a lying between 1 and 2 have been
computed* to a high degree of accuracy, so that it is possible to
find the value of F(a) for all a > 0.
It remains to define T(a) for negative values of a. The
recursion formula (81-4) can be written as
(61-5) r(«) = £<SL±1>.
The formula (81-5) becomes meaningless when a is set equal to
zero, for
lim F(a) = +00 and Km T(a) — — <*>.
a-»0-f a-»0-
It follows from (81-5) that the function F( — a) is discontinuous
when a is a positive integer.
If any number — 1 < a < 0 is substituted in the left-hand
side of (81-5), the right-hand side gives the value of r( — a);
for the values of a + 1 lie between 0 and 1, and T(a) is known
* A small table is found in B. O. Peirce, A Short Table of Integrals, p. 140.
§81 ORDINARY DIFFERENTIAL EQUATIONS
for these values of a. Thus,
rf-i'U
275
r(-o.o) = !%>, etc.
In this manner the values of T (a) for —1 < a < 0 can be com-
puted. If these values are known and the recursion formula
(81-5) is used, the values of r(«) for —2 < a < — 1 can be
obtained, etc. The adjoining figure represents the graph of
r(«) (Fig. 76).
It was observed that
r(a + !) = «!
when a is a positive integer. This
formula may serve as the definition
of factorials of fractional numbers.
Thus,
r(i) = 0! = i.
This section will be concluded
with an ingenious method of evaluating
c~xx1^ dx.
If the variable in this integral be changed by the transformation
x = ?/2, the integral becomes
,(81-6) y2\ =
Since the definite integral is independent of the variable of
integration and is a function of the limits,
(81-7) y2\ = 2
Multiplying (81-6) by (81-7) gives
(MO2 = 4 fo" e-*z*
which can be written as a double integral
(81-8) O^!)2 = 4 f " f e-tx'+Vy
»/0 J 0
-«
Ydy,
dy dz.
276 MATHEMATICS FOR ENGINEERS AND PHYSICISTS £81
In order to evaluate (81-8), transform it into polar coordinates
by setting z = r cos 0 and y = r sin 6. The element of area
dy dz becomes r dr dQ, and (81-8) becomes
= 4 I dr I r5e^2 sin2 0 cos2 0 dO.
But
ir
f* sin2 cos2 e dB = ~
Jo lo
and
/* 00
Jo e TT r = •
The latter integral is evaluated by integration by parts. There-
fore
)_ . _ /~\T* f — — . .
- 4 °r 2" ~ 2
It can be shown with the aid of the recursion formula that
It follows that (81-1) has the value t/o vW(2a) sec.
PROBLEMS
1. Compute the values of r(a) for every integer and half integer from
0 to 5 by using the relations T(l) = 1 and r(H) == V*. Plot the
curve y = T(a) with the aid of these values.
2. The Beta function B(m, n) is defined by the integral
B(m, n) = a^^-Hl - x)"~l dx.
*/o
If x is replaced by y2 in T(n) — I x*1"1^"* dx, there results
*/o
F(n) = 2 f* e-iy^dy.
•/o
Using this integral, form
T(m)r(n) = 4 f°° x*»-le~**dx f " y*"-le-*> dy.
jo «/o
Express this product as a double integral, transform to polar coordi-
nates, and show that
„<„,„> ,„<„,»,
§82
ORDINARY DIFFERENTIAL EQUATIONS
277
3. Show, by a suitable change of variable, that (81-2) reduces to
4. Show that
.cKr*
'(a) /•«
*n ~~ Jo
82. Orthogonal Trajectories. In a variety of practical investi-
gations, it is desirable to determine the equation of a family of
curves that intersect the curves
of a given family at right
angles. For example, it is
known that the lines of equal
potential, due to a distribution
of steady current flowing in a
homogeneous conducting me-
dium, intersect the lines of
current flow at right angles.
Again, the stream lines of a
steady flow of liquid intersect
the lines of equal velocity
potential (see Sec. 66) at right
angles.
Let the equation of the given family of curves be
FIG 77.
(82-1)
/O, y, c) = o,
where c is an arbitrary parameter. By specifying the values of
the parameter c, one obtains a family of curves (see solid curves
in Fig. 77). Let it be required to determine the equation of a
family of curves orthogonal to the family defined by (82-1).
The differential equation of the family of curves (82-1) can
be obtained by eliminating the parameter c from (82-1) and its
derivative,
(82.2) §f + M = 0.
dx dy ax
Let the resulting differential equation be
Now, by definition, the orthogonal family of curves cuts the
curves of the given family (82-1) at right angles. Hence, the
278 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §82
slope at any point of a curve of the orthogonal family is the nega-
tive reciprocal of the slope of the curves of the given family.
Thus, the differential equation of the desired family of curves is
This is a differential equation of the first order, and its general
solution has the form
(82-3) *(x, y, c) = 0.
The family of curves defined
by (82-3) is the desired family of
curves orthogonal to the curves of
the given family (82-1). It is
called the family of orthogonal
trajectories.
Example. Let it be required to find
the family of curves orthogonal to
the family of circles (Fig. 78)
(82-4) x> + 7/2 - ex = 0.
The differential equation of the family (82-4) can be obtained by dif-
ferentiating (82-4) with respect to x and eliminating the parameter c
between (82-4) and the equation that results from the differentiation.
The reader will check that the differential equation of the family
(82-4) is
FIG. 78.
Hence, the differential equation of the family of curves orthogonal to
(82-4) is
This is a homogeneous differential equation whose solution is easily
found to be
X2 + y2 _ cy _ 0.
Thus, the desired family of curves is the family of circles with centers
on the y-axis (see Fig. 78).
PROBLEMS
1. Find the orthogonal trajectories of the family of concentric circles
& + yz = a2.
2. Find the orthogonal trajectories of the family of hyperbolas xy = c.
ORDINARY DIFFERENTIAL EQUATIONS
279
3. Find the orthogonal trajectories of the family of curves y = cxn.
Sketch the curves of the given and the desired families for n = 1, —1,2.
4. If the equation of a family of curves is given in polar coordinates
as /(r, 6, c) — 0, show that the tangent of the angle made by the radius
vector and the tangent line at any point (r, 6} of a curve of the family is
equal to r -r- Hence, show that the differential equation of the
orthogonal trajectories of the given family of curves is obtained by
replacing r -y- by — ~ -^ in the differential equation of the given family
of curves.
5. Using the results of Prob. 4, show that the orthogonal trajectories
of the family of cardioids r — c(\ — cos 6) is another family of cardioids.
6. Find the orthogonal trajectories of the family of spirals r — ecB.
7. Find the orthogonal trajectories of the family of similar ellipses
8. Find the orthogonal trajectories of the family of parabolas
y2 = 4px.
9. Find the equation of the curve such that the area bounded by the
curve, the z-axis, and an ordinate is proportional to the ordinate.
83. Singular Solutions, It was remarked in Sec. 68 that a
differential equation may possess
solutions which cannot be obtained
from the general solution by specify-
ing the values of the arbitrary con-
stants. Such solutions are called
singular solutions.
Consider a family of integral curves
defined by
(x>y).
(83-1) <p(x, y, c) = 0,
FIG. 79.
where (83-1) is the general solution of the differential equation
(83-2)
Assume that the family of curves denned by (83-1) is such that it
has an envelope* (Fig. 79). Since the slope of the envelope at
any point (x, y) is the same as that of the integral curve which is
* It will be recalled that an envelope of a family of curves is a fixed curve C
such that every curve of the family is tangent to C.
280 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §83
tangent to the envelope at (x, y), it follows that the equation of
the envelope must satisfy (83-2). In general, the envelope is
not a curve belonging to the family of curves defined by (83-1),
and hence its equation cannot be obtained from (83-1) by
specifying the value of the arbitrary constant c. It will be
recalled that the equation of the envelope is obtained by elimi-
nating the parameter c between the equations
<t>(x, y, c) = 0
and = 0.
dc
Example. It is readily verified that the family of integral curves
associated with the equation
(83-3) i
is the family of circles
(83-4) (x - c)2 + y-
y--o
FIG. 80
The equation of the envelope of the
family (83-4) is obtained by eliminating c between (83-4) and
There results
(83-5)
-2(x - c) = 0.
y = ±
which represents the equation of a pair of lines tangent to the family of
circles (83-4) (Fig. 80). Obviously, (83-5) is a singular solution of
(83-3), for it cannot be obtained from (83-4) by any choice of the
constant c.
Inasmuch as the problem of determining the singular solutions
of a given differential equation is relatively rare in applied work,
the subject will not be pursued here any further.
REVIEW PROBLEMS
1. A particle slides down an inclined plane making an angle 0 with
the horizontal. If the initial velocity is zero and gravity is the only
force acting, what are the velocity of the particle and the distance
traveled during the time £? Compare the time of descent and the termi-
nal velocity with those of a particle falling freely from the same height
as that of the inclined plane.
§83 ORDINARY DIFFERENTIAL EQUATIONS
2. A particle falls in a liquid under the action of the force of gravity.
If the resistance to the motion is proportional to the velocity of the par-
ticle, what is the distance traveled in t seconds when the particle starts
from rest?
3. A bullet is projected upward with an initial velocity of VQ ft. per
second. If the force of gravity and a resistance that is proportional to
the velocity are the only forces acting, find the velocity at the end of
t sec. and the distance traveled by the bullet in t sec.
4. The rate of decomposition of a certain chemical substance is
proportional to the amount of the substance still unchanged. If the
amount of the substance at the end of t hr. is x and XQ is the initial
amount, show that x = Zoe-*', where k is the constant of propor-
tionality. What is the constant of proportionality if x changes from
1000 g. to 500 g. in 2 hr.?
6. A torpedo moving in still water is retarded with a force propor-
tional to the velocity. Find the speed at the end of t sec. and the dis-
tance traveled in t sec., if the initial speed is 30 miles per hour.
6. A disk is rotating about a vertical axis in an oil bath. If the
retardation due to friction of the oil is proportional to the angular
velocity o>, find w after t sec. The initial velocity is w0.
7. Water is flowing out through a circular hole in the side of a
cylindrical tank 2 ft. in diameter. The velocity of the water in the
jet is \/*2gh, where h is the height in feet of the surface of the water
above the center of the orifice. How long will it take the water to fall
from a height of 25 ft. to a height of 9 ft. above the orifice, if the orifice
is 1 in. in diameter?
8. Water is flowing out from a 2-in. horizontal pipe running full.
Find the discharge in cubic feet pejr second if the jet of water strikes the
ground 4 ft. beyond the end of the pipe when the pipe is 2 ft. above
the ground.
9. A projectile is fired, with an initial velocity ?>o, at an angle a with
the horizontal. Find the equation of the path under the assumption
that the force of gravity is the only force acting on the projectile.
10. A cylindrical tumbler containing liquid is rotated with a con-
stant angular velocity about the axis of the tumbler. Show that the
surface of the liquid assumes the shape of a paraboloid of revolution.
Hint: The resultant force acting on a particle of the liquid is directed
normally to the surface. This resultant is compounded of the force of
gravity and the centrifugal force.
11. Two chemical substances combine in such a way as to produce
a compound. If the rate of combination is proportional to the product
of the unconverted amounts of the parent substances, find the amount
of the compound produced at the end of time L The initial amounts of
the parent substances are a and 6.
282 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §83
Hint: dx/dt = k(a - x)(b - x).
12. Assume that the pressure p of the air at any height h is equal to
the weight of the vertical column of air above it. If the density of the
air is proportional to the pressure, what is the law connecting the
pressure p with the height hi
13. A particle of mass m is sliding down a rough inclined plane (the
coefficient of friction /x = 0.2), whose height is 300 ft. and whose angle
of inclination is 30°. If the particle starts from rest, how long will it
take to reach the foot of the plane? With what velocity will it be
traveling then?
Hint: The differential equation of
motion is
d*s ,• ^
jp = 0(sm a - AI cos a),
where a is the angle of inclination of
e plane.
14. A runaway carrier in an aerial
1000 the plane.
_,
r 1G. ol .
tramway is moving along the arc of a
second-degree parabola joining the points whose coordinates are (0,
0) and (1000, 300) (Fig. 81). How long will it take the carrier to
reach the lowest point if the factional resistance is neglected and if the
carrier starts from the top with initial velocity zero? See in this con-
nection the Engineers1 Bulletin of the Colorado Society of Engineers,
June, 1935.
16. A brick is set moving in a straight line over the ice with an
initial velocity of 20 ft. per second. If the coefficient of friction between
the brick and the ice is 0.2, how long will it be before the brick stops?
16. A certain radioactive salt decomposes at a rate proportional to
the amount present at any instant t. How much of the salt will be
left 300 years hence, if 100 mg. that was set aside 50 years ago has been
reduced to 90 mg.?
17. A skier weighing 150 Ib. is coasting down a 10° incline. If the
force of friction opposing the motion is 5 Ib. and the air resistance is
two times the speed in feet per second, what is the skier's speed after
t sec.?
18. A tank contains 1000 gal. of brine holding 1 Ib. of salt per gallon.
If salt water containing 2 Ib. of salt per gallon is allowed to enter the
tank at the rate of 1 gal. per minute and the mixture, which is kept
uniform by stirring, is permitted to flow out at the same rate, what is
the amount of salt in the tank at any time £?
Hint: Let the amount of salt present at any time t be x\ then, the rate
at which x changes is equal to the rate of gain, in pounds per minute,
§84 ORDINARY DIFFERENTIAL EQUATIONS 283
diminished by the rate of loss. Thus,
dx x
dt 1000
19. A 100-gal. tank contains pure water. If 50 per cent alcohol is
allowed to enter the tank at the rate of 2 gal. per minute and the
mixture of alcohol and water, which is kept uniform by stirring, leaves
the tank at the same rate, what is the amount of alcohol in the tank
at the end of 10 min.?
20. The rate at which two chemical substances are combining is
proportional to the amount of the first substance remaining unchanged.
If initially there are 20 Ib. of this substance and 2 hr. later there are only
10 Ib., how much of the substance will be left at the end of 4 hr.?
21. A series circuit consists of a condenser whose capacity is c farads
and the resistance is 72 ohms. Before the circuit was closed the con-
denser contained a charge of #0 coulombs. What is the charge on the
condenser t sec. later? (The differential equation is R -,7 + - = 0.)
dt c
22. The rate at which a body is cooling is proportional to the differ-
ence in the temperatures of the body and the surrounding medium.
It is known that the temperature of a body fell from 120° to 70°C. in
1 hr., when it was placed in air at 20°C. How long will it take the
body to cool to 40°C.? 30°C.? 20°C.?
23. A bullet is fired vertically down from a balloon that is 2 miles
above the surface of the earth. On the assumption that the resistance
is proportional to the square of the velocity, find the velocity with
which the bullet strikes the earth if the initial velocity is 1800 ft. per
second.
84. Linear Differential Equations. The remainder of this
chapter will be restricted to the treatment of linear differential
equations, that is, equations of the type
(84-1) Po(*)^ + ?!(*) |3+ +P»-i(*)fc + P*Wv=fM,
where the Pi(x) and/(z) are either functions of x or constants. It
is extremely fortunate that a large number of physical phe-
nomena are successfully described with the aid of linear differ-
ential equations. It will be shown in the succeeding sections
that it is possible to give a more detailed account of the treat-
ment and solution of linear differential equations than has been
furnished for non-linear equations.
284 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §86
85. Linear Equations of the First Order. A linear differential
equation of the first order has the form
In order to solve this equation, set y = uv, where u and v are
functions of x that are to be determined later. With this
substitution, (85-1) becomes
dv . du . * , N - , N
Udx + V~dx+fl(x^UV~f*^
or
(85-2)
If u is suitably chosen, the bracket in (85-2) can be made equal
to zero, thus reducing (85-2) to a simple form. In order to
choose u so that the expression in the bracket is equal to zero, set
du
or
~
It
Integrating gives
log u + J/i(x) dx = c,
and choosing the simplest expression for u, by setting c = 0,
produces
With this choice of u, (85-2) becomes
-r/jC*) dx dv _
dx ~~
or
— _ f/i(») dxf
dx "~ e J
which integrates into
§86 ORDINARY DIFFERENTIAL EQUATIONS 285
By hypothesis, y = uv, so that
(85-3) y = e-W> dx / e/''(a° d%(x) dx + ce~I'>(x) dx.
This is the general solution of (85-1).
Example 1. Solve
-7- + y cos x = sin 2x.
Upon using formula (85-3),
y = e-I COB x r** f e/cos * dx sin 2# eta + ce~f°°a * dx
= e~«n * J esia x sin 2x dx + ce~«™ x,
which is easily evaluated by replacing sin 2x by 2 sin x cos x.
Example 2. Solve
Here,
r2<JE
^ = 6~Jx+i /
^ ±J ™^ (x + D2
wliich is easily evaluated.
PROBLEMS
Solve the following equations*
1. (1 + x') dy + (xy - £) <fa - 0.
2. (a;2 + 1) fx + 2xy = x*.
3. = ^' - 2xy.
6. ^ — h y cos a: = cos3 x.
6. x ^ + y - x2 sin x = 0.
dy _ y - 1
'' dc x* + 1
8. L -r + El = j&, given that 7 = 0 when J = 0; L, ^, and E are
constants.
286 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §86
dy
9. T- = y + cos x — sin x.
10. -r — y sec x esc x = e*(l — sec x esc a?).
11. -i — \- yx = y.
12. dx + 2xdy — y dy = 0.
13. ^ + ?/ sec2 # = tan z sec2 x.
14. (x + 1) ~/x - y = e* (x + I)2.
15. ~T~ — 2?/ — 63* = 0.
86. A Non-linear Equation Reducible to Linear Form (Ber-
noulli's Equation). An equation of the type
(86-1) -~ + fi(x)y = fz(x)yn,
in which n may be regarded different from zero and unity, can
be reduced to linear form by the substitution z = yl~n. Then,
dz ,1 . dy
-r- = (1 - ri)y~n —•
dx '* dx
and (86-1) becomes
g- (n - !)/!(*)* = -(n-
which is a linear equation in z.
Example. Solve
Setting z = I/yz, the equation becomes
whose general solution is
z
so that
§87 ORDINARY DIFFERENTIAL EQUATIONS 287
PROBLEMS
Solve the following equations:
2. -
3 + _=a;2
7/6 dx xy5
~ ~l '
5. x -^ + y = 2/2 log
6. + xy = xV.
_ -
cte 1 - a;2 ~ 1 - a;2'
87. Linear Dififerential Equations of the nth Order. No
formulas are available for the solution of the linear differential
equation, with variable coefficients, of order greater than 1.
This section contains some interpretations of the symbolic
notation that will be found useful in the solution of the linear
differential equation
in which the al are constants.
It will be convenient to introduce the new notation
%-Dv and m D»y.
dx y dxn y
In this notation, (87-1) becomes
Dny + ai Dn~ly + a* Dn~*y + - • • + an-i Dy + any = f(x)
or
(87-2) (D» + oiZ)-1 + a2D-2 + • • • + an-iD + a»)y = /(x).
The expression in the parentheses in (87-2) is known as a linear
differential operator of order n. Obviously, it is not an algebraic
288 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §87
expression multiplying y but is a symbol signifying that certain
operations of differentiation are to be performed on the function y.
Thus, D2 — 2Z) + 5 operating on log x gives
(D2 - 2D + 5) log x s D2 log x - 2D log x + 5 log x
_ d2 log x ~ d log # ,r i
1 2 , „ ,
The gain in simplicity in using the operational notation results
from the fact that linear differential operators with constant
coefficients formally obey the laws which are valid for poly-
nomials. Thus
so that the operator D is distributive. If the symbol
(D + ai)(D + at),
where ai and o2 are constants, is interpreted to mean that the
operator D + ai is applied to (D + a^)y, then
' (D + oi)(D + a,)y = (D
d
(ai + a2) Dy
= [D2 + (ai + a2)Z) +
It is readily established that operating on t/ with
(D + a2)(D + ai)
produces precisely the same result. Hence, the commutative
law holds, or
(D + oi)(D + a2)2/ s (D + a,)(D
s [D2 + (ax + a2)D +
It is readily established that the law of exponents also holds,
namely,
§87 ORDINARY DIFFERENTIAL EQUATIONS 289
so that linear operators can be multiplied like ordinary algebraic
quantities, where the powers of D in the result are to be inter-
preted as successive differentiations.
The solution of (87-2) can be written in the symbolic form
1 ,, ,
y D» + a,D^ + • • • + an-iD + anJ(X)'
The meaning of this symbol will be investigated next.
Consider a simple differential equation
(87-3) ;! = /(*) or A/ =/(*)•
The solution of (87-3), in symbolic form, is
so that the symbol l/D must be interpreted as integration* with
respect to x. Thus,
1 ,, x
The meaning of a more complicated symbol can be obtained
from a consideration of the first-order equation
(87-4) g + ay = f(x),
where a is a constant. Writing this equation in the operational
notation, it becomes
(D + a)y = /(*).
The symbolic solution in this case is
(87-5) y - -f(x).
It was established in Sec. 85 that the general solution of
(87-4) is
(87-6) y = ce~ax + e~ax J eaxf(x) dx,
* In order to make the definition of the operator l/D unambiguous, one
could agree that the constant of integration should be selected so that
y = 0 when x assumes some specific value. However, in order to avoid
complication, the constant that arises from the integration of f(x) will be
suppressed.
290 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §87
and it is desirable to give the symbolic solution (87-5) an inter-
pretation that is consistent with the actual solution (87-6).
Now, the solution (87-6) consists of two parts, the first of which,
ce~ax, if taken alone, obviously does not satisfy (87-4). The
second part
e-ax J caxf(x) dx
is a solution of (87-4), for (87-6) represents the general solution
which reduces to
e-ax J ea*f(X} fa
when the arbitrary constant is taken as zero. The part of the
solution (87-6) containing f(x) is called a particular integral of
(87-4), and the part containing the arbitrary constant is called
the complementary function. It may be observed that the
complementary function cc~ax satisfies the homogeneous linear
differential equation*
It is convenient to associate with the symbol (87-5) the particular
integral of (87-4), namely,
f(x) ES e— J
(87-7) ^~ f(x) ES e— e?*f(x) dx.
The arbitrary constant arising from the integration in (87-7) may
be taken as zero, for the addition of this constant of integration-
will give rise to a term that can be merged with the complemen-
tary function. The integral operator
as defined by (87-7), is of fundamental importance in the follow-
ing sections. The meaning of a more complicated symbolic solu-
tion will be given later.
Example 1. To interpret the symbol
1
D + a x">
* The term homogeneous linear differential equation should not be confused
with the homogeneous equation discussed in Sec. 76. The homogeneous
linear differential equation is one of the type (84-1), where f(x) B 0.
§88 ORDINARY DIFFERENTIAL EQUATIONS 291
write out its meaning with the aid of (87-7). Then,
i*xm dx
m~l m(m — l)xm~2
.f . _
, if m £ 0,
except when a = 0. If a = 0 and m ^ — 1, then
Example 2.
— /v»m
Dx
c . a sin mx — m cos mx
sm mx = e~a:c I eaa: sin mo: do? =
J
^r—. — = = - ^—. -
D + a J a2 + m
PROBLEMS
1. Show that
2. What is the meaning of jr—. — emx<!
U ~f~ ft
3. What is the meaning of yr— r — cos mx?
D ~\~ a
88. Some General Theorems. In Sec. 87, it was found that
the general solution of the non-homogeneous linear differential
equation of the first order contained as part of itself the solution
of the homogeneous equation
It will be shown next that a similar statement can be made
concerning the general solution of the nth-order linear equation.
Consider first a homogeneous linear differential equation of the
nth order with constant coefficients,
If y = emx is substituted in this equation, the result is
(mn + aimn~l + • • • + an_ira + an)emx = 0.
If m is chosen so that it satisfies the equation
(88-2) mn + cum*-1 + • • • + an-im + an = 0,
292 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §88
which is called the auxiliary, or characteristic, equation, then
y = emx will be a solution of (88-1). But (88-2) has, in general,
n distinct roots, Wi, w2, • • • , mn, so that there will be n distinct
solutions
Because of the linear character of (88-1), it is clear that, if y = emf
is a solution, then
y = ctew/,
where ct is an arbitrary constant, is also a solution. Moreover,
it is readily verified that the sum of the solutions of a homogeneous
linear differential equation is also a solution of the equation.
Thus,
(88-3) y = ciems + c2ems + - - • + cnemnx
will be a solution; and since it contains n arbitrary constants (all
roots rat are assumed to be distinct), it is the general solution of
(88-1).
Let
(88-4> +*+ •••+"-+«*-/<'>.
where f(x) ^ 0 ; and assume that, by inspection or otherwise, a
solution y = u(x) of (88-4) has been found. Then, if (88-3)
is the general solution of the homogeneous equation (88-1),
(88-5) y = Ciemix + c^em^x + • • • + cnem*x + u(x)
will be the general solution of (88-4). This fact can be verified by
direct substitution of (88-5). That (88-5) is the general solution
follows from the fact that it contains n arbitrary constants
The part of (88-5) that is denoted by u(x) is called a particular
integral of (88-4), and the part containing the arbitrary constants
is called the complementary function.
Example 1. Solve
dx* dx* dx ~~
The auxiliary equation is
m8 — rn2 — 2m = 0,
and its roots are mi = 0, ra2 = — 1, m3 = 2. Then the complementary
§88 ORDINARY DIFFERENTIAL EQUATIONS 293
function is
Y = d + C&-* + c&**.
A particular integral u(x) is
u(x) = Hxe~x.
Therefore, the general solution is given by
y=Y + u(x).
If (88-1) is written in symbolic form as
(88-6) (D- + aJJ*-1 + • • • + an^D + an)y = 0
and the differential operator (which is of precisely the same form
as the auxiliary equation defined above) is treated as an algebraic
expression, then (88-6) can be written as
(88-7) (D - mi)(D - m2) • • • (D - mn)y = 0.
Consider the n first-order linear homogeneous equations
(D - mi)y = 0,
(D - m,)y = 0,
(D - mjy = 0,
whose solutions can be obtained at once by recalling that the
meaning of the symbol is given by
(D - m)y ss -| - my.
These solutions are emix, emzx, • • • , em«x, which are precisely
the same as the solutions obtained for (88-1) by a different
method.
The general solution of (88-7) was found to be (88-3) under
the assumption that all the roots rat were distinct. If some of the
roots are equal, the number of arbitrary constants ct in (88-3)
is less than n and the solution given there is not the general
solution. Suppose that the equation
is such that its auxiliary equation has a double root, that is,
mi = W2 = m. Then this equation can be written as
(D - m)(D - m)y = 0. ,
294 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §88
If (D — m)y is set equal to v, the equation becomes
(D - m)v = 0
and v = Ciemx is its solution. Since (D — w)t/ = v, it follows that
(D - m)y = ciemx,
which is a linear equation whose solution can be found, with the
aid of (87-6), to be
y = emx(c2
Thus, if the auxiliary equation has a double root, the solution
corresponding to that root is
y = emx(Ci + CiZ).
By an entirely similar argument, it can be established that, if the
auxiliary equation possesses a root m of multiplicity r, then the
solution corresponding to that root is
y = emx(Ci + C2Z + ' ' ' + CrX1^1).
Example 2. Find the solution of
(D3 - 3D2 + 4)y = 0.
The auxiliary equation is
m3 - 3m2 + 4 = 0 or (m + l)(m - 2)2 = 0.
Therefore the general solution is
Example 3. Find the solution of
(D2 + l)y - 0.
The auxiliary equation is
m2 + 1 = 0 or (m - f)(m + i) = 0.
Therefore, the general solution is
y = de»* + c2e-»* = A cos 3 + B sin a?.
PROBLEMS
1. Find the general solutions of
§89 ORDINARY DIFFERENTIAL EQUATIONS 295
»>§-«!+«.-».
(d) (D3 — 2Z)2 + D)y = 0.
(e) (D4 + 3D3 + 3D2 + D)y = 0.
(/) (Z)4 — Jb4)y = 0.
to) (D3 — 3D2 + 4)y = 0.
(h) (D3 - 13D + 12)y = 0.
(f) (D3 + D2 - D + l)y = 0.
(j) (D* + 2D3 + D*)y = 0.
}. The Meaning of the Operator
1
D* + aiD^-1 + • • • + an~iD + an
f(x).
In Sec. 87 the meaning of the operator n — /(x) was given.
AX ™| G/
Now, consider a second-order linear differential equation with
constant coefficients,
3 +<••!+ «*-"*>
or
(89-1) (Z>2 + oiD + a,)y = /(x).
It was remarked in Sec. 87 that linear operators with constant
coefficients obey the ordinary laws of algebra and can be treated
as polynomials. Therefore, (89-1) can be factored to read
(D -m1)(D - ro,)» = f(x)
or
(D - mOy = j^M
= emix f e~mixf(x') dx,
in accordance with (87-7). Hence,
(89-2) y = j^ ems \ <r"i*J(x) dx
JJ — niz J
B em** \ e^i-™*** I e~mi*f(x) dx dx.
296 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §89
For mi = w2, (89-2) reduces to
(89-3) y = emix J J e~mixf(x) dx dx.
By direct substitution in (89-1), it is easy to establish the fact
that (89-2) is a particular solution of (89-1). The general
solution, according to Sec. 88, is made up of the sum of (89-2) and
the general solution of the homogeneous equation
3 +*£+—*
which is known to be
y = ciemi* + C2em2*, mi ^ m2,
or
y = (ci + Czx)emi*, mi = m2.
The interpretation of the symbol
which represents the symbolic solution of the differential equa-
tion
dny . dn~ly . dy . ,, .
d£ + <»d^+ ••• +an^/x + any=:f(x),
or
(89-4) (D» + axD"-1 + • • • + an_,Z) + o»)y = /(*),
can now be made easily. Write the operator in (89-4) in factored
form,
• • • + an_x + an
= (D - mi)(D - m,) • • • (D - mn),
so that (89-4) becomes
V = (D - mn}(D - mn_0 • • • (D -
1 1
_ _
D - TOn D - mn_i D - m/>
Successive operations on /(z) with -~ - give
jLx ~~ tn/i
(89-5) y = emix J e(w2~wi)x f e(mrmJx • • • J e~mnxf(x) (dx)n>
and the result is a particular integral of (89-4).
§89 ORDINARY DIFFERENTIAL EQUATIONS 297
It can be shown that if the operator
_ 1 _
D" + axD*-1 + • - • + an_iZ) + an
is decomposed into partial fractions (the denominator being
treated as a polynomial in D), then
_ _ 1 _ ff v
y » ^1 + - • • + an^D + anJ(X)
Al 4- Az An
+
vD - mi D - m2 ^ ^ D - mn
which gives, by (87-7),
(89-6) y = Aiemix J e~mixf(x) dx + A2em2* J e~mff(x) dx
which is also a particular integral of (89-4).
Thus, there are available two methods for the determination
of the particular integral. The first method* of finding the
particular integral, (89-5), is known as the method of iteration,
and the second, (89-6), as the method of partial fractions.
Generally speaking, formula (89-6) is easier to apply. However,
if the roots of the auxiliary equation are not all distinct, the
decomposition of the operator into partial fractions, of the type
considered, cannot be effected and formula (89-5) must be used.
Example 1. Solve
d*y dy
-'-'•- — 5 i — h G?/ — 6
dx2 dx J
or
(Z)2 - 5D + 6)7; = e**
or
The particular integral, as obtained by the method of iteration, is
11 1
/>x —
e ~
r
I /
J
- - — ,
D -3 D -2 ~ D -
= eZx C (e~* C e** dx\ dx = y-
298 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §89
If the method of partial fractions is used, then
„ 1 1_ ,4, _ /_J L_\ ,4,
y~D-3D-2e \D-3 D-2)e
f f e*x
= e** j e **e**dx - e** J e **e** x = -y
The complementary function is
therefore, the general solution is
Example 2. Solve
or
The particular integral is
= x - 2,
and the general solution is given by
y = (ci + c2x)e-* + x — 2.
PROBLEMS
1. Solve jjjj + 3j/ = a;'.
2.
4. The flexure y for end thrust P is given by
where E is Young's modulus, w is the load, and I is the moment of
inertia. Solve this equation.
5. Solve (D3 - 2D2 - D + 2)y = 1 - 2*.
6. Solve (Z)2 + >££> - H)0 = cos x - 3 sin x.
7. Solve (D3 - 3D + 2)y = 2 sin * - 4 cos x.
8. Solve (D2 - l)y = 5x - 2.
9. Solve (D3 - D2 - 8D + 12)y - 1.
10. Solve (D4 — l)y == e* cos x.
11. Solve (D* - 2D + l)y - «e».
§90 ORDINARY DIFFERENTIAL EQUATIONS 299
12. Solve (Z)2 + D - 2)y = sin 2z.
13. The differential equation of the deflection y of the truss of a
suspension bridge has the form
where H is the horizontal tension in the cable under dead load q, h is
the tension due to the live load p, E is Young's modulus, and I is the
moment of inertia of the cross section of the truss about Xhe horizontal
axis of the truss through the center of gravity of the section and per-
pendicular to the direction of the length of the tru&=. Solve this
equation under the assumption that p — qh/H is a constant.
14. The differential equation of the deflection y of a rotating shaft
has the form
where El is the flexural rigidity of the shaft, p is the mass per unit
length of the shaft, and o> is the angular velocity of rotation. Solve
this equation.
15. The differential equation of the buckling of an elastically sup-
ported beam under an axial load P has the form
£!V_,jP^_L*-n
dx* + El dx* + El y ~ U>
where El is the flexural rigidity and k is the modulus of the foundation.
Solve this equation.
90. Oscillation of a Spring and Discharge of a Condenser.
The foregoing discussion gives all the essential facts for solving
an nth-order linear differential equation with constant coeffi-
cients. At this point, it is desirable to apply the methods of
solution, outlined above, to a group of important practical
problems.
Suppose that it is required to determine the position of the end
of a helical spring at any time t. It is assumed that the spring is
set vibrating in a vacuum so that considerations of damping do
not enter here. If a mass M (Fig. 82) is applied to the end of
the spring, it produces an elongation s which, according to
Hooke's law, is proportional to the applied force. Thus,
F - ks,
where F = Mg from the second law of motion and k represents
300 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §90
the stiffness of the spring. Then,
Mg = ks.
If at any later time t an additional force is applied to produce an
extension y, after which this additional force is removed, the
spring will start oscillating. The problem is to determine the
position of the end point of the spring at any subsequent time.
The forces acting on the mass M are the force of gravity Mg
downward, which will be taken as the positive direction for
the displacement y, and the tension T in the
spring, which acts in the direction opposite
to that of the force of gravity. Hence, from
Newton's second law of motion,
k(s+y) Since T is the tension in the spring when its
elongation is s + y, Hooke's law states that
T = k(s + y), so that
FIG. 82.
But Mg — ks, and therefore the foregoing
equation becomes
Setting k/M — a1 reduces this to
(90-1)
dt*
+ a*y = 0 or (D2 + a2)*/ = 0.
Factoring gives (D — ai)(D + ai)y = 0, from which it is clear
that the general solution is
Recalling that etx = cos x + i sin x (Sec. 73), the solution can
be written as
y = Ci(cos at — i sin at) + C2(cos at + i sin a£)
= A cos at + B sin at,
where A = ci + c2 and B = (c2 — Ci)i. The arbitrary con-
§90 ORDINARY DIFFERENTIAL EQUATIONS 301
stants A and B can be determined from the initial conditions.
The solution reveals the fact that the spring vibrates with a
simple harmonic motion whose period is
The period depends on the stiffness of the spring as would be
expected — the stiffer the spring, the greater the frequency of
vibration.
It is instructive to compare the solution just obtained with
that of the corresponding electrical problem. It will be seen
that a striking analogy exists between the mechanical and
electrical systems. This analogy is responsible for many recent
improvements in the design of telephone equipment.
Let a condenser (Fig. 83) be discharged through an inductive
coil of negligible resistance. It is known that c
the charge Q on a condenser plate is proportional _.
to the potential difference of the plates, that is,
Q = CV, L
where C is the capacity of the condenser. FIG. 83.
Moreover, the current 7 flowing through the coil is
_ _dQ
~ ~dt'
and, if the inductance be denoted by L, the e.m.f. opposing V is
L dl/dt, since the IR drop is assumed to be negligible. Thus,
or
C~J
Simplifying gives
dzQ I n __ n
-575 + 77f v = 0,
ill ' \jJU
which is of precisely the same form as (90-1), where a2 = 1/CL,
and the general solution is then
302 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §91
The period of oscillation is
T = 27r VOL.
Note that the inductance L corresponds to the mass M of the
mechanical example and that 1/C corresponds to the stiffness k
of the spring.
91. Viscous Damping. Let the spring of the mechanical
example of Sec. 90 be placed in a resisting medium in which the
damping force is proportional to the velocity. This kind of
damping is termed viscous damping.
Since the resisting medium opposes the displacement, the
damping force r -~ acts in the direction opposite to that of the
d/t
displacement of the mass M . The force equation, in this case,
becomes
or, since Mg = fcs,
^j-.L^ + Atf-o
dt* ^ Mdt ^ M y ~ u*
In order to solve this equation, write it in the more convenient
form
(91-D -g+2(> !+„.„,„.
In this case the auxiliary equation is
m2 + 26m + a2 = 0
and its roots are
m = -6 ± \A2 - a2,
so that the general solution is
(91-2) y = cie^-^^^1^^ + c2e(-fe- V**^)'.
It will be instructive to interpret the physical significance of
the solution (91-2) corresponding to the three distinct cases that
arise when fc2 - a2 > 0, fc2 - a2 = 0, and b* - a2 < 0. If
62 — a2 is positive, the roots of the auxiliary equation are real
and distinct. Denote them by mi and m^ so that (91-2) is
(91-3) y = ciemi' + c2e"V.
§91
ORDINARY DIFFERENTIAL EQUATIONS
303
The arbitrary constants Ci and c2 are determined from the initial
conditions. Thus, let the spring be stretched so that y = d and
then released without giving the mass M an initial velocity.
The conditions are then
y = d
when t = 0 and
when t = 0.
Substituting these values into (91-3) and the derivative of
(91-3) gives the two equations
d = Ci + c2 and 0 =
These determine
m^d ,
Ci = — — — and
+ m2c2.
mid
mi — m2
Hence, the solution of (91-3) is
mi — m2
The graph of the displacement represented as a function of t is of
the type shown in Fig. 84.
Theoretically, y never becomes
zero, although it comes arbi-
trarily close to it. This is the
so-called overdamped case.
The retarding force is so great
in this case that no vibration
can occur.
If b2 — a2 = 0, the two roots
of the auxiliary equation are equal and the general solution of
(91-1) becomes
y =
FIG. 84.
If the initial conditions are
when t = 0 and
y = d
0
dy
dt
304 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §91
When t = 0, the solution is
y = de~bt(l + bf).
This type of motion of the spring is called dead-beat. If the
retarding force is decreased by an arbitrarily small amount, the
motion will become oscillatory.
The most interesting case occurs when b2 < a2, so that the
roots of the auxiliary equation are imaginary. Denote 62 — a2
by —a2, so that
m = — 6 ± ia
and
y = Cie(-b+i<x)t _|_ C2e(-6-u«)l
= e~bt(A cos at + B sin at).
If the initial conditions are chosen as before,
y = d
when t = 0 and
when £ = 0, the arbitrary constants A and B can be evaluated.
The result is
y = de~bt f cos a£ H — sin a£ J,
which can be put in a more convenient form by the use of the
identity
A cos 6 + B sin 6 = V?2 + B* cos (0 - tan-1 ~Y
The solution then appears as
(91-4) y = - V^+T2 <r-6< cos
a
( orf - tan-1 -\
\ a/
The nature of the motion as described by (91-4) is seen from
Fig. 85. It is an oscillatory motion with the amplitude decreasing
exponentially. The period of the motion is T = 2ir/a. In the
undamped case the period is T = 27r/a; and since
§91 ORDINARY DIFFERENTIAL EQUATIONS 305
it follows that
27T 2*
a. d
Thus the period of oscillation is increased by the damping.
An electrical problem corresponding to the example of the
viscous damping of a spring is the following: A condenser (Fig.
86) of capacity C is discharged through an inductive coil whose
resistance is not negligible. Referring to Sec. 90 and remember-
\ ^-^
C
—
^
»_
VWWVWW-
R
FIG. 85.
L
FIG. 86.
ing that the IR drop is not negligible, the voltage equation is
found to be
V -L%- 7/2 = 0
at
or
Simplifying gives
W r L ~dt """ CL ~~ v'
and this equation is of the same form as that in the mechanical
example. The mass corresponds to the inductance L, r corre-
sponds to the electrical resistance Ry and the stiffness k corresponds
to 1/C. Its solution is the same as that of the corresponding
mechanical example and is obtained by setting 26 = R/L and
a2 = l/CL.
PROBLEMS
1. The force of 1000 dynes will stretch a spring 1 cm. A mass of
100 g, is suspended at the end of the spring and set vibrating. Find the
306 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §91
equation of motion and the frequency of vibration if the mass is pulled
down 2 cm. and then released. What will be the solution if the mass is
projected down from rest with a velocity of 10 cm. per second?
2. Two equal masses are suspended at the end of an elastic spring
of stiffness k. One mass falls off. Describe the motion of the remain-
ing mass.
3. The force of 98,000 dynes extends a spring 2 cm. A mass of
200 g. is suspended at the end, and the spring is pulled do\vn 10 cm. and
released. Find the position of the mass at any instant t, if the resistance
of the medium is neglected.
4. Solve Prob. 3 under the assumption that the spring is viscously
damped. It is given that the resistance is 2000 dynes for a velocity
of 1 cm. per second. What must the resistance be in order that the
motion be a dead beat?
6. A condenser of capacity 4 microfarads is charged so that the
potential difference of the plates is 100 volts. The condenser is then
discharged through a coil of resistance 500 ohms and inductance 0.5
henry. Find the potential difference at any later time t. How large
must the resistance be in order that the discharge just fails to be
oscillatory? Determine the potential difference for this case. Note
that the equation in this case is
d*V dV V
6. Solve Prob. 5 if R = 100 ohms, C = 0.5 microfarad, and L =
0.001 henry.
7. A simple pendulum of length / is oscillating through a small
angle 0 in a medium in which the resistance is proportional to the
velocity. Show that the differential equation of the motion is
Discuss the motion, and show that the period is 2w -y/co2 — k* where
co2 = g/l
8. An iceboat weighing 500 Ib. is driven by a wind that exerts a
force of 25 Ib. Five pounds of this force are expended in overcoming
frictional resistance. What speed will this boat acquire at the end
of 30 sec. if it starts from rest?
Hint: The force producing the motion is F = (25 — 5)0 = 20gr.
Hence, 500 dv/dt = 200.
9. A body is set sliding down an inclined plane with an initial
velocity of v0 ft. per second. If the angle made by the plane with the
horizontal is 6 and the coefficient of friction is ju, show that the distance
§91
ORDINARY DIFFERENTIAL EQUATIONS
307
traveled in t sec. is
6 —
, cos
Hint: m d*s/dt* = mg sin 6 — nmg cos 6.
10. One end of an elastic rubber band is fastened at a point P, and
the other end supports a mass of 10 Ib. When the mass is suspended
freely, its weight doubles the length of the band. If the original length
of the band is 1 ft. and the weight is dropped from the point P, how far
will the band extend? What is the equation of motion? w
11. It is shown in books on strength of materials and
elasticity that a long beam lying on an elastic base, the
reaction of which is proportional to the deflection y,
satisfies the differential equation
Set a4 = k/(±EI), and show that the characteristic
equation corresponding to the resulting differential equa-
tion is m4 + 4a4 = 0, whose roots are m = ± a + ai.
Thus show that the general solution is
y = Cieax cos ax +
sin ax + c$e~ax cos ax
sn ax.
FIG. 87.
12. If a long column is subjected to an axial load P and the assump-
tion that the curvature is small is not made, then the Bernoulli-Euler
law gives (see Sec. 72)
dx*
M_
El'
Since the moment M is equal to — Py (Fig. 87), it follows upon setting
dy/dx — p that the differential equation of the deformed central axis is
D^
dy Py
(1 + pi)fc == ~ Ji'
Solve this differential equation for p, and show that the length of the
central line is given by the formula
where k2 = d*P/4EI, d is the maximum deflection, and F(k, tr/2) is
the elliptic integral of the first kind. The equation of the elastic
curve, in this case, cannot be expressed in terms of the elementary func-
tions, for the formula for y leads to an elliptic integral.
308 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §92
92. Forced Vibrations. In the discussion of Sec. 91, it was
supposed that the vibrations were free. Thus, in the case of the
mechanical example, it was assumed that the point of support
of the spring was stationary and, in the electrical example, that
there was no source of e.m.f. placed in series with the coil.
Now, suppose that the point of support of the spring is vibrating
in accordance with some law which gives the displacement of the
top of the spring as a function of the time t, say x — /(£), where x
is measured positively downward. Just as before, the spring
is supposed to be supporting a mass M, which produces an
elongation s of the spring. If the
displacement of the mass M from its
position of rest is y, it is clear that,
when the top of the spring is dis-
placed through a distance x, the
vwwwwwwmooooomooo J a°tual extension of the spring is
R
tion is
R L y — x. If the resistance of the
medium is neglected, the force equa-
Jtf «f = Mg - k(s + y - x) = -k(y - x),
whereas, if the spring is viscously damped, it is
Upon simplifying this last equation, it becomes
(92-1) M^ + r^ + ky^kx,
where x is supposed to be a known function of t.
The corresponding electrical example is that of a condenser
(Fig. 88) placed in series with the source of e.m.f. and that dis-
charges through a coil containing inductance and resistance.
The voltage equation is
where f(f) is the impressed e.m.f. given as a function of t. Since
_r-dQ _ dV
1 ~ dt ~ C -&'
§92 ORDINARY DIFFERENTIAL EQUATIONS 309
the equation becomes
d~V dV
(92-2) CL+CR + V
An interesting case arises when the impressed e.m.f. is sinusoidal,
for example,
f(f) = EQ sin cot.
Then the equation takes the form
dW RdV 1 I,-
Both (92-1) and (92-2) are non-homogeneous linear equations
with constant coefficients of the type
»
(92-3) ^ + 26^ + a«y = o»/(0.
The solution of this equation is the sum of the complementary
function and the particular integral (see Sec. 88). The com-
plementary function has the form shown by (91-2), namely
c\emJ + c2em2*,
where
mi = —b + \/fr2 — a2 and ra2 = —b — \/b2 — a2.
The particular integral, by (89-5), is
(92-4) Y = aW J e^~m^ [ J er<*Sf(t) dt] dt.
From the discussion of Sec. 91, it is clear that the part of the
solution which is due to free vibrations is a decreasing function
of t and will become negligibly small after sufficient time has
elapsed. Thus the " steady-state solution" is given by the
particular integral (92-4).
Let it be assumed that the impressed force, x in (92-1) and
f(t) in (92-2), is simply harmonic of period 27r/co and of amplitude
a0. Then,
f(f) = a0 sin ut,
and (92-4) becomes
Y = aW J e(mrm^ (J e-"Va0 sin wt dt) dt.
310 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §93
The result of integration* is
a2a0
Y = — = sin (wt — c),
- co2)2 + 46V
where
€ = tan"1
This is the steady-state solution.
The remainder of this section will be devoted to the physical
interpretation of the solution (92-5). It is observed that if the
impressed frequency is very high (large w), then the amplitude
of the sinusoid (92-5) is small, so that the effect of the impressed
force is small. When co = a, the amplitude is a0a/26, which may
be dangerously large if b (and hence the resistance of the medium)
is small. For a fixed 6 (resistance of the medium) and a (natural
frequency of the system), the maximum amplitude occurs when
(a2 — w2)2 + 4b2co2 is a minimum, that is, when
-^ [(a2 - co2)2 + 4b2co2] = 0.
do)
This is readily found to be when
co2 = a2 - 262.
Upon recalling the physical significance of a and 6, these results
can be interpreted immediately in terms of the physical quantities.
93. Resonance. It was remarked in Sec. 92 that if the
impressed frequency is equal to the natural frequency of vibra-
tion, then the amplitude of (92-5) may be abnormally large.
Stated in terms of the physical quantities of the electrical and
mechanical examples, this means that the maximum voltage
of the electrical system may be dangerously large or that the
maximum displacement of the spring may be so great as to pro-
duce rupture.
The phenomenon of forced vibration is of profound importance
in many engineering problems. Not so many years ago the
collapse of a building in one of the larger American cities was
* Integration in this case is a little tedious. For actual integration, it is
convenient to replace sin ut by the equivalent exponential expression
§93 ORDINARY DIFFERENTIAL EQUATIONS 311
attributed to the rhythmic swaying of the dancing couples, who
happened to strike the natural frequency of the beam supporting
the structure. Again, the failure of the Tacoma bridge was
explained by some on the basis of forced vibration. It is also
well known that soldiers are commanded to break step in crossing
a bridge, for fear that they may strike the note of the cables.
The walls of Jericho are reported to have fallen after seven priests
with seven trumpets blew a long blast.
The phenomenon of resonance occurs when the impressed
frequency is equal to the natural frequency. Consider Eq.
(92-3) in which 6 (resistance) is zero and f(t) = a0 sin at} so that
(93-1) ~ + a*y = a2a0 sin at.
The particular integral in this case is
(93-2) F = a0a2e-a" J (e2a < J e~^ sin at dt) dt,
since mi = ai and ra2 = —ai. If sin at is replaced by ^ >
Zi
(93-2) integrates into
V — 2 (t COS a^ I 1 I * 4\
JL — ~~"~ a$a \ ^ ~i~ o Q sin ai ~\ ^ „ cos ai j.
If F is added to the complementary function c\ cos at + c2 sin at,
the general solution is given by
(93-3) y = A cos at + B sin at ~- t cos at,
where the last two terms of F have been combined with the
complementary function. Let the initial conditions be y = 0
when t = 0, and dy/dt = 0 when t = 0. Then A = 0 and
B = a0/2, and (93-3) will be
(93-4) y = -~ (sin at — at cos a£).
This equation represents a vibration whose amplitude increases
with time; for the amplitude of the first term is the constant
a0/2, and the amplitude of the second term is proportional to the
time t. In fact, if sufficient time is allowed, the amplitude may
312 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §94
become greater than any preassigned number. This remark
ought not to stimulate the student to design an apparatus to
produce an infinite amplification or an infinite force. In any
physical case, there is some resistance b present, and a brief
reference to (92-5) will show that b prevents the oscillations from
becoming arbitrarily large.
PROBLEMS
Show that a particular integral of
1 d*y . 2 • , • 1 4
!• 37s + a y — sin at 1S y = ~~ o~ t cos at*
at 4a
2. -jif + azy — cos at is y = -~-t sin at.
at *j(i
94. Simultaneous Differential Equations. In many investi-
gations, it is necessary to consider systems of differential equa-
tions involving several dependent variables and one independent
variable. For example, the motion of a particle in the plane can
be described with the aid of the variables x and y, representing
the coordinates of the particle, each of which may depend on
time. It will be indicated in this section how a system of n
ordinary differential equations involving n dependent variables
may be reduced to a study of a single differential equation of
higher order.
Let two dependent variables x and y be functions of an inde-
pendent variable t, and let it be required to determine x and y
from the simultaneous equations
dx
(94-1) dt
= /2(0,
where a, b, c, and d are constants. If these equations are written
in operational form, they are
(D + a)x + by = /i(Z),
ex + (D + d}y = /2(0-
Operating on the second of these equations with - (D + a) gives
c
(D + a)x + - (D + a)(D + d)y = - (D
C C
§94 ORDINARY DIFFERENTIAL EQUATIONS 313
and, if the first equation is subtracted from this result,
i (D + a) CD + d)y - by = \ (D + a)/8(0 - fi(t).
c c
This is a second-order linear differential equation which can be
solved for y. In order to determine x, solve the second equation
of (94-1) for x,
-t-*}
and substitute the value of y in terms of t.
The reader may show in the same way that the solution of a
system of two second-order linear differential equations can be
reduced to the solution of a linear differential equation of the
fourth order (see Example 2 below). /////////
Example 1. Consider ^
J 4- 2x - 2y = t,
dy _
dt ~" X y ~~ e
or
(D + 2)x - 2y = t,
-3* + (D + l)y = 6'.
Operate on the second of these equations with
+ 2) to obtain
-(Z) + 2)x + H(D + 2)(D + l)y = y3(D + 2)e',
and add this result to the first equation. The result is
y3(D + 2)(D + l)y - 2y = H(D + 2)e< + t,
which simplifies to
This equation can be solved for y as a function of t, and the result can
be substituted in the second of the given equations to obtain x.
Example 2. Let the two masses MI and M 2 be suspended from two
springs, as indicated in Fig. 89, and assume that the coefficients of
stiffness of the springs are k\ and k% respectively. Denote the dis-
placements of the masses from their positions of equilibrium by x
and y. Then it can be established that the following equations must
314 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §94
hold:
These equations can be simplified to read
d*y , kt k% ^
w+w^-ws-0'
dzx kz . ki + kz
W -Mly + —MTx = "-
By setting
ki kz Mz
M\ = a ' Wz = b > Mi = m'
the equations reduce to
(D2 + 62)?/ - b*x = 0,
-62/ni/ + (D2 + a2 + 62m)x = 0.
Operating on the second of these reduced equations with 7-7- (D2 + 62)
and adding the result to the first of the equations give
(D2 + 62)(Z)2 + a2 + b*m)x - b*mx = 0
or
[D4 + (a2 + &2 + 62m)Z)2 + aV]x = 0.
This is a fourth-order differential equation which can be solved for
# as a function of t. It is readily checked that
x = A sin (<ut — e)
is a solution, provided that w is suitably chosen. There will be two
positive values of co which will satisfy the conditions. The motion of the
spring is a combination of two simple harmonic motions of different
frequencies.
PROBLEMS
1. Solve Examples 1 and 2, Sec. 94.
2. The equations of motion of a particle of mass m are
where a?, y, z are the coordinates of the particle and X, Y, Z are the
components of force in the directions of the #-, T/-, and 2-axes, respec-
tively. If the particle moves in the zy-plane under a central attractive
§95 ORDINARY DIFFERENTIAL EQUATIONS 315
force, proportional to the distance of the particle from the origin, find
the differential equations of motion of the particle.
3. Find the equation of the path of a particle whose coordinates x
and y satisfy the differential equations
d^x dy
(Py __ dx
where H, E, e, and m are constants. Assume that x = y = dx/dt
= dy/dt — 0 when t = 0. This system of differential equations occurs
in the determination of the ratio of the charge to the mass of an electron.
4. The currents /i and 72 in the two
coupled circuits shown in Fig. 90 satisfy the - - - - ™ ->-*
following differential equations:
d2/! d*I2 d/2 , /2
M ~2 + L — 4-r — + — - 0
Reduce the solution of this system to that of a single fourth-order dif-
ferential equation. Solve the resulting equation under the assumption
that the resistances Ri and R% are negligible.
95. Linear Equations with Variable Coefficients. With the
exception of linear equations with constant coefficients and such
equations with variable coefficients as are reducible to those
with constant coefficients by a change of variable, there are no
general methods for solving linear differential equations of order
higher than the first. In general, solutions of differential
equations with variable coefficients cannot be expressed in terms
of a finite number of elementary functions, and it was seen in a
number of specific examples that the solutions of such equations
lead to new functions which are defined either by definite integrals
or by infinite series. Some of these functions are of such frequent
occurrence in applied mathematics that it has been expedient to
calculate their values and tabulate them, precisely as the values
of logarithms and trigonometric functions are tabulated. It
must be borne in mind that the term elementary function as
applied to logarithmic and circular functions is, in a sense, a
misnomer and that such functions as Gamma functions, Bessel
functions, and Legendre polynomials become just as "elemen-
tary" after their values have been tabulated. The elementari-
316 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §96
ness of any given function is measured by the ease with which its
value can be ascertained.
The remainder of this chapter contains a brief treatment of
those linear differential equations which are of common occur-
rence in practice. An attempt will be made to express the
solutions in convergent power series in x. This involves the
tacit assumption that the solutions are capable of being expanded
in Maclaurin's series, which, of course, is not true in general, and
it is therefore not surprising that occasionally this method fails
to give a solution. The method consists in assuming that a
solution of the differential equation
is expressible in a convergent infinite series in powers of #, of the
type
(95-2) y = a0 + a& + a2x2 + • • • + anxn + - • • ,
where the coefficients at are to be determined so that the series
will satisfy the differential equation. If the coefficients of the
derivatives in (95-1) arc polynomials in x, then the obvious mode
of procedure is to substitute the infinite scries (95-2) into the
equation (95-1), expand f(x) in Maclaurin's series, combine the
like powers of a*, and equate to zero the coefficient of each power
of x. This leads to an infinite set of algebraic equations in the
at, which can sometimes be determined by algebraic means.
It is stated without proof that a homogeneous linear differ-
ential equation of order n,
(95-3) g + Pl(*) g^ + • • • + ?_!<*) | + Pn(x)y = 0,
where the pt are continuous one-valued functions of x, possesses
n linearly independent solutions, and only n. If these solutions
are yi(x), yi(x), • • • , yn(x), then the general solution of the
equation is given by
(95-4) y = ciyi + c22/2 + • • • + cnyn.
This fact can be immediately verified by substituting (95-4) in
(95-3). It is also clear that, if u(x) is any particular solution of
§96
ORDINARY DIFFERENTIAL EQUATIONS
317
(95-1), then its general solution is y — c\y\ + c2?_/2 + * * *
+ cnyn + u(x), where Cit/i + c2t/2 + • • • + cnyn is the solution
of the related homogeneous equation (95-3).
Frequently, it is of practical importance to knqw whether a
given set of functions is linearly independent. Inasmuch as
the definition for linear independence that is given in Sec. 34 is
difficult to apply, a test for the linear independence of the solu-
tions will be stated.*
THEOREM. The necessary and sufficient condition that a given
set of solutions y\, y%, • • • , yn of the nth order differential equation
(95-3) be linearly independent is that the determinant
2/1
W 3
2/2
2/2'
- - - ynr
This determinant is called the Wronskian.
Example. By substitution, it can be verified that y\ = sin x,
7/2 = cos x, and 7/3 = elx are solutions of the differential equation
But the Wronskian is
cj dx* ^ dx u
sin x cos £ eix
cos x —sin £ ie13
—sin x —cos £ — etx
= o,
and therefore this set of solutions is not linearly independent. In other
words, at least one of them can be expressed as a linear combination of
the other two. It is known that
el* — cos x + i sin x.
It is readily verified that a linearly independent set of solutions is
T/i = sin x, 7/2 = cos x, and 7/3 = ex, so that the general solution is
y = ci sin x + c2 cos x + c3ex.
•
PROBLEMS
Determine whether or not the following sets of functions are linearly
independent:
* See INCB, E. L., Ordinary Differential Equations.
318 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §96
1. 2/1 = sin x + x, 2/2 == ext 2/3 ~ 3e* — 2x — 2 sin x.
2. 2/1 = x2 - 2x + 5, 2/2 = 3z - 7, 2/3 = sin x.
3. 2/1 = 6** + x, 2/2 = cos a: + x, yz = sin x.
4. 2/1 = (z + I)2, 2/2 = (x - I)2, 1/3 = 3*.
5. 2/1 = log x, 2/2 = sinh x, 2/3 = e*, 2/4 = e~
~*
96. Variation of Parameters. Two methods of determining a
particular integral of a linear differential equation with constant
coefficients were discussed and illustrated in Sec. 89. Another
important method that is applicable to linear equations with
either constant or variable coefficients will be described here.
This method, due to the great French mathematician Lagrange
(1736-1813), permits one to determine a particular integral of
(96-1) g + Pl(x) g3 + • • • + Pn-^x) I + p.(x)y = /(*),
when the general solution of the related homogeneous equation
' ' '
(96"2) n- Tx
is known.
Let the general solution of (96-2) be
(96-3) y = Ciyi + c2yt + • • • + cnyn,
in which the cl are arbitrary constants, and assume that a set of
n functions #1(0;), vi(x), • • • , vn(x) can be so chosen that
(96-4) y = viyi + v^y^ + • • • + vnyn
will satisfy (96-1). Since y\(x), y<t(x), • • • , yn(x) are known
functions of x, (96-1) imposes only one condition upon the vl
in (96-4). Inasmuch as there are n functions vt, it is clear that
n — 1 further independent conditions can be imposed upon the
vt, provided that these conditions are consistent.
Differentiating (96-4) gives
' ' • + vny'n) + (v{yi + vfM* + • • • + v'nyn).
As one condition to be imposed on the v^} let
v(y\ + v'2y2 + • • • + v'nyn = 0,
so that
§96 ORDINARY DIFFERENTIAL EQUATIONS
Then,
319
and if the second condition to be satisfied by the v% is
• • • + <y'« = o,
it follows that
V" = viy[' + v&y
+ vny"n.
By continuing this process a set of n — 1 conditions is imposed
on the Vt, namely,
- • • • + v'nyn = 0,
h • • • + <//: = o, .
(96-5)
as a consequence of which
yf =
• • • + vnyn,
i(n-l)
Calculating y(n} yields
Substituting t/, yf, • • • , y(n) in (96-1) and remembering that,
by hypothesis, y\, t/2, • • • , yn satisfy (96-2) give the nth con-
dition to be imposed upon the vt, namely,
The n — 1 relations (96-5) together with (96-6) give n linear
algebraic equations which can be solved for vj, v2> ' * ' , v'n, pro-
vided that the determinant of the coefficients of the v(, namely,
(i 2/2 ' ' ' yn
9l 2/2 ' ' ' 2/n
is not identically zero. But this determinant is the Wronskian
and, since yi, 2/2, • • • , yn were assumed to be linearly inde-
320 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §96
pendent, it is different from zero. Hence, the system of equations
can always be solved for the v't, and the expressions for the ut are
obtained by integration.
Example. As an illustration* of the application of this method of
determining a particular integral, consider the equation
2» = 2(sinx-2coB*).
The general solution of the homogeneous equation is found to be
Assume that a particular integral of the non-homogeneous equation is
of the form
y — v\ex
where Vi, v^, and vz are functions of x to be determined presently.
Computing yr gives
y1 = vtfx + v2(x + \}e* — 2v3e~'2x + Vi'e* + v2xex + v3fe~2x.
The first condition to be imposed upon the v, is
Vi'e' + v2'xe* + v3'e~2* = 0, (1)
so that
y" * Vl€* + V2(x + 2)e* + 4vse~Zx + vi'e* + v2'(x + \)e* - 2v^e~2x
Imposing the second condition produces
vi'e9 + v2'(x + l)ex - 2v3'e~** = 0, (2)
and computing y1" yields
y'" = viex + vt(x + 3)6* - Svse-** -f vi'e* + v>2'(x -f 2)ex -f 4vz'e~*x.
Hence, the third condition to be satisfied by the vt is
vi'e* + vj(x + 2)e* + 4z>3V2* = 2(sin x - 2 cos a). (3)
Solving (1), (2), and (3) for Vi, v2f, and vs' gives
Vi' = — ?^6"*(sin a: — 2 cos 3) — %e~x(sm ^ — 2 cos x),
iV - ^^"'(sin x — 2 cos x),
Vs = %e2a!(sin a; — 2 cos x).
The integration! of these expressions yields
* For another illustration, see Example, Sec. 97.
t The integration in this case is quite tedious, and generally speaking it
is easier to solve linear equations with constant coefficients by the methods
§96 ORDINARY DIFFERENTIAL EQUATIONS 321
Vi = Hxe~x(3 sin x — cos x) + e~* sin x + %e~* cos x,
t>2 = }ie~x( — 3 sin a; + cos a;),
t>3 = — %e2x cos x,
in which the constants of integration are omitted because a particular
integral is desired.
By hypothesis, a particular integral is given by
y = Vie* + v2xex + v^e~Zx — sin x,
so that the general solution of the non-homogeneous equation is
y = (c\ + c>2x)e* + cze~2x + sin x.
PROBLEMS
1. Solve Probs. 1, 2, and 3, Sec. 89, by the method of variation of
parameters.
2. Find the solution of
by the method of variation of parameters, and compare your result with
that of Sec. 85. The solution of the related homogeneous equation is
obtained easily by separation of the variables.
3. By the method of variation of parameters, find a particular integral
of
d*y 3dy 5
where the general solution of the related homogeneous equation is
i , K
y = ~ + czx6.
4. Find the general solution of
&y , x dy __ \_ .
dx* "*" 1 - x dx 1 - x y ~ -1 *'
where the general solution of the related homogeneous equation is
c\ex + cix.
5. Find the general solution of
_ x*y" - 2xy' + 2y = x log x, _
discussed in Sec. 89. However, the method of the present section is of
great value when the given equation has variable coefficients,
322 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §97
if the general solution of the related homogeneous equation is
y = cix2 + C&.
97. The Euler Equation.* Before proceeding to illustrate the
method of solution in terms of infinite series, it will be well to
discuss one type of differential equation with variable coefficients
that can be reduced by a change of variable to a differential
equation with constant coefficients.
Consider the linear equation
(97-1) *• g + a^-> g3 + • • • + «„_,* | + any = /(*),
where the at are constants. This equation can be transformed
into one with constant coefficients by setting x = ez. For if
x = ez, then
dx , dz
•j- — ez and -j- = e~*.
dz dx
Moreover, if D s= --, then
9 dz
dx dz dx
and
Similarly,
g = e-3* (£>3 _ 3£
Then, since x = ez, it follows that
x ^ - TDjy
X dx~Dy'
x* = (D* - D)y = D(D
x"n = D(D - 1)(D - 2) • • • (D - n
* Also called Cauchy's equation,
§97 ORDINARY DIFFERENTIAL EQUATIONS 323
so that (97-1) is replaced by an equation with constant coefficients,
[D(D - 1) • • • (D - n + 1) + (nD(D - 1) • • • (D - n + 2)
+ • • - + an_!Z) + an]y = /(*).
Example. Consider
d*y dy
x W + xdx-y = xlo*x'
Upon making the substitution x = ee, this equation becomes
or
(D3 - 3D2 + 3D - l)y = ze*.
The roots of the auxiliary equation are MI = w2 = m^ = 1, so that the
complementary function is
The particular integral is
so that, in terms of z, the general solution is
and, in terms of x,
y — [ci + c2 log x + c3(log x)2]x -
A particular integral for this example will be obtained by the method
of variation of parameters in order to demonstrate the applicability
of this method to equations with variable coefficients. Care must be
taken first to transform the equation so that it has unity for its leading
coefficient, for the discussion of Sec. 96 was carried through for this
type of equation.
Expressing the given equation in the form (96-1) gives
d*y , 1 dy 1 1 .
Since the general solution of the homogeneous equation was found
to be
CiX + C& log X + C&(\0g X)*,
324 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §97
the equations of condition (96-5) and (96-6) are
Vi'x + v2'x log x + Va'x(log x)2 «= 0,
vi' + fi'U + log x) + vz'[(log xY + 2 log x] = 0,
v*'\ + v* (x log x + x) =
Solving for vi, t>2', and i;3' yields
Vl' ** 2x (log *)*' v* = ~~x (log ^2> v* = 2x log ^J
which integrate into
vl = H(log xY, v2 = -K(log x)3,
Hence, a particular integral is
1 /I NO
y = v& + v& log a; + y3x(log a;)2 =
PROBLEMS
1. Find the general solution of Prob. 3, Sec. 96, by the method of
Sec. 97.
2. Find the general solution of
Compute the particular integral by the method of variation of
parameters.
3. Solve
by assuming a solution of the form y — xr and determining appropriate
values of r.
4. Solve
by assuming a solution to be of the form y = xr.
5. Find the general solution of
«
x*y'" - 4zV + 5xy' - 2y - 1.
6. Find the general solution of
y oc X*.
§98 ORDINARY DIFFERENTIAL EQUATIONS 325
7. Find the general solution of
x*y" - 2xy' + 2y = x log x.
98. Solution in Series. Many differential equations occurring
in applied mathematics cannot be solved with the aid of the
methods described in the preceding sections, and it is natural to
attempt to seek a solution in the form of an infinite power series.
The method of solution of differential equations with the aid of
infinite series is of great importance in both pure and applied
mathematics, and there is a vast literature on the subject. This
section and the four following sections contain only a brief
introduction to the formal procedure used in obtaining such
solutions.
As an illustration of the method, consider the differential
equation
(98-1) y' - xy - x = 1,
and assume that it is possible to obtain the solution of (98-1) in
the form of a convergent power series
(98-2) y = a0 + a+x + a2z2 + - - - + anxn + • • • .
Inasmuch as the series of derivatives of a convergent power series
is convergent, one can write
(98-3) y' = ai + 2a2z + • - - + nanxn~l + • • • .
Substituting (98-2) and (98-3) in (98-1) and collecting the coeffi-
cients of like powers of x give
(98-4) a! + (2a2 - a0 - l)x + (3o8 - ajx* + • • •
+ (nan - an-z)xn-1 +•••=!.
By hypothesis, (98-2) is a solution of (98-1), and therefore*
equating the coefficients of like powers of x in (98-4) leads to the
following system of equations :
a i — I (coefficient of x°),
2a2 — a0 — 1 = 0 (coefficient of x),
3a3 — a\ = 0 (coefficient of a:2),
(98-5)
nan — an-2 = 0 (coefficient of a;"*1),
* See Theorem 5, Sec. 10.
326 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §98
The system of equations (98-5) is a system of linear equations
in infinitely many unknowns a0, ai, • • • , an, • • • . Solving
the second equation of (98-5) for a2 in terms of a0 gives
The third equation taken with the first demands that
aa = = .
Setting n = 4 in the coefficient of xn~l gives
i -| [ -|
&2 do ~t~ J- &0 i J-
4 2*4 2^ • 2^'
whereas n = 5 gives
„ _«3_ 1
^° 5 3-5
In general,
(98-6)
2n 2*-n!
1
(2n + 1)
The substitution in (98-2) of the values of ak given by (98-6)
leads to a solution in the form
. , OQ + 1 o , 1 n i ttO + 1 A , 1 c ,
= --1 »4 t ' ' '
When the terms containing a0 are collected, there results
+ 1 o , tto + 1 A ,
(98-7) y = ao +
+ 2n-nT X n +
I*
TU+-]-
1-3-5 •••
If ao + 1 is set equal to c, one can write (98-7) in the more
§98 ORDINARY DIFFERENTIAL EQUATIONS 327
compact form
(98-8) j,-e[i+*! + 5£n + ... +-=_, .+
g2n
2^2! ^ ^ 2^T!
f x3 x2n+l ' 1
+ [~~l+X + T^+ ' ' ' + 1 - 3 • • • (2n + 1) + ' ' '}
The two series appearing in (98-8) are easily shown to be
convergent for all values of x, and hence they define functions
of x. In fact, the first of the series is recognized as the Maclaurin
expansion of e2, so that (98-8) can be written as
^ T rr3
(98-9) y = ce2 + 1-1 + x + ~-^ + • • •
T2n-fl
+ 1-3 . . . (2n+l) +
}
This is the general solution of (98-1), for it contains one arbitrary
constant.
Since (98-1) is a linear differential equation of the first order,
its solution could have been obtained by using the formula
(85-3), and it is readily verified that (85-3) gives
X2
(98-10) y = ce2 - 1 + e2 J e 2 dx.
The integral in (98-10) cannot be evaluated in closed form; but
if the integrand is expanded in a power series in x, it is easy to
show that (98-10) leads to (98-9).
Consider next the homogeneous linear differential equation of
the second order
(98-11) y" - xy' + y = 0,
and assume that the solution of (98-11) has the form
(98-12) y = anxn = a0 + a& + - • • + anxn + - • •
n=X)
Then the series for y' and y" are
00 00
i'' = V nanX71"1 and y" = V n(n — l)anzn~2.
n>-2
328 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §98
If these expressions are substituted in (98-11), the result is
n(n - l)ana:n*"2 - nanxn + anxn = 0.
n—2 n-\ n = 0
Combining the terms in like powers of x and setting the coeffi-
cient of each power of x equal to zero give the system of equations
2 • 1«2 + flo = 0 (coefficient of x°),
3 • 2a3 — ai + ai = 0 (coefficient of x),
4 • 3a4 — 2a2 + a2 = 0 (coefficient of z2),
............................................... >
(n + 2)(n + l)an+2 — nan + an = 0 (coefficient of xn),
Hence,
(98-13) an+2 =
This recursion formula can be used to determine the coefficients
in (98-12) in terms of a0 and a\. Thus, substituting n = 0,
1, 2, • • • in (98-13) gives
1
=- - o0,
1 1
T^[ a2 = ~ 4]
2
4-75
3 3
6!
a2n+i = 0,
1 • 3 • 5 • • • (2n - 3)
a2n = --- __ -
Therefore,
(98-14) y =
where a0 and a\ are arbitrary. It is readily checked that the
§99 ORDINARY DIFFERENTIAL EQUATIONS 329
series is convergent, and, since it is a power series, it defines a
continuous function of x.
The two linearly independent solutions of (98-11) are then
y =
and
1 2 1 4 3 , 15 8
-
>
These solutions are obviously linearly independent, for one of
them defines an odd function of x and the other defines an even
function of x.
PROBLEMS
Integrate in series:
3. (x* -3» + 2) + (x»-2i-l)+(*- 3)y = 0.
4- ft - y - 1.
cfc2 ^
99. Existence of Power Series Solutions. It must be kept
clearly in mind that the calculations performed in Sec. 98 are
formal and depend on the assumption that the differential equa-
tions discussed there possess power series solutions. If, for
example, an attempt had been made to apply the method of
solution outlined in Sec. 98 to the equation
xy' - 1 = 0,
it would have been futile. The general solution of this equation
is
y = log x + c,
which cannot be expanded in a power series in x.
The task of determining beforehand whether a given differen-
tial equation possesses solutions in the form of power series
represents one of the major problems of analysis. It will suffice
in this introductory treatment to state, without proof, the
conditions under which a homogeneous linear differential equa-
tion of the second order has a power series solution.
330 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §99
THEOREM. Let
(99-1) V" + fi(x)y' + f*(x)y = 0
have coefficients f\(x) and f%(x) that can be expanded in power series
in x which converge for all values of x in the interval —R < x < R ;
then there will exist two linearly independent solutions of the form
00
y = 2} anXn>
n»0
which will converge for all values of x in the interval —R < x < R.
It is clear from the statement of the theorem that the differen-
tial equation will possess power series solutions, which converge
for all values of x, whenever f\(x) and/2(x) are polynomials in x.
It should be noted that the coefficient of the second derivative
term in (99-1) is unity. Frequently, the differential equation
has the form
and if this differential equation is put in the form (99-1), then
/. / x Pi(x) j
fi(x) - and
If po(x) should vanish for some value of x in the interval within
which the solution is desired, one must expect trouble. If
po(x), pi(x), and pz(x) are polynomials in x and if 7>o(0) ^ 0, then
one can surely expand fi(x) and fa(x) in power series in some
interval, and the theorem enunciated above is applicable.
As an illustration, consider the differential equation
(99-2) (2 - x)y" + (x - l)y' - y = 0. "
Inasmuch as
and
/2(x) = (x - 2)-1
obviously possess power series expansions that are valid in the
interval — 2 < x < 2, it is reasonable to proceed to obtain the
power series solution.
§99 ORDINARY DIFFERENTIAL EQUATIONS 331
Substituting
y =
in (99-2) gives
(2 - x) n(n - l)anx»-* + (x - 1) wa^"-1 - J anz" = 0.
n=0 n=0 n-0
Rearranging and combining terms give
n-O
(n -
" = 0.
Equating the coefficients of the powers of x to zero gives
(99-3)
2 • 2 • Ia2 — oi — a0 = 0,
2 • 3 • 2a3 - 22a2 = 0,
2 • 4 • 3a4 - 32a3 + a2 = 0,
2(w + 2)(n + l)an+2 - (w + l)2an+1 + (n - l)an = 0,
The coefficient of xn provides the recursion formula
2(n + l)(n + 2)
and setting n = 0, 1, 2, 3, • • • gives
#o ~f" di c
22 ~ 2 - 2!
c
,» where c = a0 +
a4 =
3 2-3!
32a3 — a2 c
4! 2-4!
42a4 - 2a3 c
"• 2-4-5 2-5!
It is easily shown that in general
— °
a" "" 2^T!?
332 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §100
so that
y = a0 + a\x + a2x2 + • • • + anxw + • • •
= ao + ^i^ +
/a:2 x*
\2i + n
2
+ *!/, , , *2
^0 + &\X +
^& \ ^1
flo H~ ^i
(1
2 2
= Cie* + c2(l - x),
where Ci = (a0 + ai)/2 and c2 = (a0 — ai)/2.
It happens in this illustration that the series appearing in the
solution of the differential equation represent elementary func-
tions, so that the general solution can be written in* closed form
Ordinarily, the infinite series arising from the solution of linear
differential equations with variable coefficients represent func-
tions that carniQt be expressed in terms of a finite number of
elementary functions. This is the case discussed in the next
section, where the series that provides the solution of the differen-
tial equation represents a class of functions of primary importance
in a great many problems in applied mathematics.
It is quite obvious that the theorem of this section can be
rephrased to include the case where the f unctions fi(x) and/2'(^)
possess series expansions in powers of x — a. In this case,
there will exist two linearly independent solutions of the form
y =
PROBLEM
Solve
y" - (x - 2)7/ = 0
00
by assuming the solution in the form y — S an(x — 2)n. Also,
n = 0
00
obtain the solution in the form y = 2 anxn.
n=0
^ 100. Bessel's Equation. An important differential equation
was encountered by a distinguished German astronomer and
§100 ORDINARY DIFFERENTIAL EQUATIONS 333
mathematician, F. W. Bessel (1784-1846), in a study of planetary
motion. The range of applications of the functions arising from
the solution of this equation is partly indicated by the fact that
these functions are indispensable in the study of vibration of
chains, propagation of electric currents in cylindrical conductors,
problems dealing with the flow of heat in cylinders, vibration of
circular membranes, and in many other problems arising in every
branch of applied mathematics. Some of the uses of this equation
are indicated in the chapter on Partial Differential Equations.
BesseFs equation has the form
(100-1) x*y" + xyf + (x2 - n2)y = 0,
where n is a constant. It will be observed that this equation
does not satisfy the conditions of the theorem of Sec. 99 because
of the appearance of x2 in the coefficient of y". In the notation
of Sec. 99,
1 n2
fi(x) = - and /2(x) = 1 - — ,
JU JU
and it is clear that /i(x) and f*(x) cannot be expanded in power
series in x.
In order to solve (100-1), assume that the solution can be
obtained in the form of a generalized power series, namely,
(100-2) y = xm 2 arXr,
where m is a constant to be determined later and where ao can be
assumed to be distinct from zero because of the indeterminate
nature of m.
Calculating the first and second derivatives with the aid
of (100-2) and forming the terms entering into (100-1) give
d2y
x2 ~ = m(m — I)a0£m + m(m + l)dixm+1 + • • •
CLX
+ ak(m + k)(m + k — l)xm+k + • • • ,
dt/
~- Q + (m + l)ai$m+l + ' • • + ak(m
x2y = aQx
— n'2y = ~-
334 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §100
Adding the left-hand members of these expressions and equating
the result to zero give (100-1). By hypothesis, (100-2) is a
solution, and therefore the coefficient of each power of x in the
sum must vanish. Hence,
(w2 — n2)a0 = 0 (coefficient of xm),
[(m + I)2 - n2]ai = 0 (coefficient of xm+l),
,
[(m + fc)2 - n2]a,c + a,t_2 = 0 (coefficient of xm+k).
The coefficient of the general term gives the recursion formula
(100-3) a, = - r-^l-l 5-
^ / (m + k)2 — n2
The equation resulting from equating to zero the coefficient of the
term of lowest degree in x (here, xm) is known as the indicial
equation. In order to satisfy the indicial equation, choose
m = ±n. It m is chosen as -\-n or —n, aQ is arbitrary and the
second condition requires that «i = 0.
For m = n the recursion formula becomes
* (n + fc)2 - n2 fc(2n + fc)
Setting fc = 2, 3, 4, • • * in turn gives
a2 = -
^^rt -j- ^;
Ch
since ai =
2(2n + 2)'
PI
3 3(2n + 3) =
a«>
a4 = -
4(2n + 4) 2 • 4(2n + 2)(2n + 4)
^3 = 0
6 5(2n + 5) '
In this manner as many coefficients as desired can be computed;
and if their values in terms of a0 are substituted in (100-2), there
is obtained the following series, which converges for all values of x}
(100-4) y = atf
2) ' 2-4(2n + 2)(2n + 4)
x6
2 • 4 • 6(2n + 2)(2n + 4)(2n + 6)
§100 ORDINARY DIFFERENTIAL EQUATIONS 335
If m = — n is chosen, ao is again arbitrary and a\ = 0, and
the resulting series is
«„*-» [
(100-5) » = «„*-» l + — +
— 274(2X2^4)
X6
2 • 4 • 6(2n - 2)(2n - 4)(2n - 6)
The series (100-4) and (100-5) become identical for n = 0. For*
positive integral values of n, (100-5) is meaningless, since some of
the denominators of the coefficients become zero, and (100-4) is
the only solution obtainable by this method. If n is a negative
integer, (100-4) is meaningless and (100-5) is the only solution in
power series in x. For n 7^ 0 or an integer, both (100-4) and
(100-5) are convergent and represent two distinct solutions.
Then (100-4) multiplied by an arbitrary constant and added to
(100-5) multiplied by an arbitrary constant gives the general solu-
tion of the Bessel equation.
The reason for the failure of this method to produce two dis-
tinct solutions when n is zero or an integer is not hard to find.,
The success of this method depends upon the assumption that the
solutions are representable in power series. The analysis leading
to the determination of the second particular solution of (100-1)
when n is an integer is not given here. * It is sufficient to mention
the fact that the second solution can be obtained by assuming
that it has the form, when n is a positive integer,
oo
(100-6) i/2 = Cyi(x) log x + V akx-n+k,
*=o
where y\(x) is the solution (100-4) and C is a constant. Obvi-
ously, this solution becomes infinite when x = 0.
It will be of interest to consider the particular solution obtained
from (100-4) by setting
a° =
*See WATSON, G. N., Theory of Bessel Functions; GRAY, A., G. B.
MATHEWS, and J. M. MACROBEBT, A Treatise on Bessel Functions and
Their Applications to Physics; WHITTAKER, E. J., and G. N. WATSON,
Modern Analysis; MCL.ACHLAN, N. W., Bessel Functions for Engineers.
336 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §100
The series (100-4) then becomes
/y»n /fW+2 /yn-f-4
(100-7) Jn(x) m ^ • 2n+2(*n + 1}! + ^(n + 2)!
_ . . . + (_l)fc ^n+2fc -4- -..
~ v A/ 2n+2*A!l^ti. 4- fc")!
;(-!)»-
-2
The function defined by this series is called the Bessel function
of the first kind of order n. This series holds for non-negative
values of n. For n = 0,* (100-7) gives
x2 * zk
JQ(x) = 1 - ^2
and for n = 1,
23 • 2! 252!3!
i)!
which are called Bessel functions of the zero-th and first orders,
respectively.
The formula (100-7) can be generalized for non-integral
positive values of n with the aid of the Gamma function by
writing (n + k)\ = T(n + k + 1), so that
(100-8) ^(')
For n = Y<i, (100-8) becomes
/2fc + 8\'
For n « 0, n! is defined to be unity. This is consistent with the formula
when n = 1, as well as with the general definition of n! in Sec. 81.
§100 ORDINARY DIFFERENTIAL EQUATIONS
It is not difficult to show that this reduces to*
sin x.
337
TTX
This formula suggests that the behavior of Bessel functions may
be somewhat similar to that of the trigonometric functions. This
proves to be the case, and it will be shown in the next section that
Bessel functions can be used to represent suitably restricted
arbitrary functions in a way similar to that in which trigonometric
functions are used in Fourier series.
* It is clear that JQ(X) is an even function and that Ji(x) is an
odd function of x. Their graphs are shown in Fig. 91. For large
FIG. 91.
values of x the roots of JQ(X) = 0 and J\(x) = 0 are spaced
approximately TT units apart. It can be shown that for large
values of x the Values of Jn(x) are given approximately by the
formula
It is not difficult to show with the aid of (100-6), by setting
yi(x) = Jo(x), that the second solution of BessePs equation of
order zero has the form
= J,(x) log x + -
l + +
22.42
(2fc)2
This function is called the Bessel function of the zero-th order
* See Prob. 2 at the end of this section.
338 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §100
and the second kind. Thus, the general solution of the equation
is
y = CiJoGO + CzKQ(x),
where C\ and Cz are arbitrary constants.
Bessel functions of the first kind and of negative integral order
are defined* by the relation
/_„(*) = (-1)V«(*).
The values of Bessel functions are tabulated in many books, f
Electrical engineers frequently use the real and imaginary parts
of Jn(\/—ix)J which are denoted by the symbols bernx and
beinx, respectively, that is,
~—~i x) = bernx + i beinx.
There are also modified Bessel functions of the first kind, which
are denoted in the literature by In(x). The commonly used
notations for modified Bessel functions of the second kind are
7n(x), Nn(x), Hn(x).
PROBLEMS
1. Show that
^;/o(z) = -Ji(x).
2. Show that
Note that T(n + 1) = nT(n) and
3. Show that y = JQ(kx) is a solution of the differential equation
xy" + y' + k*xy = 0.
4. Show that
x*y" + xy' + (kW - n*)y = 0, k ^ 0,
can be reduced to the form
* See BYERLY, W. E., Fourier Series and Spherical Harmonics, Sec. 120,
p. 219.
t JAHNKE-EMDE, Funktionentaf cln ; BYERLY, W. E., Fourier Series and
Spherical Harmonics; WATSON, G. N., Theory of Bessol Functions.
§101 ORDINARY DIFFERENTIAL EQUATIONS 339
Hint: Set z = kx, and hence y' = -r- -r- = A; -7-
^-
/2~
6. Show that «/_!,$ (2) = ^J — cos x, so that the general solution of
\/ 7T3J
Bessel's equation with n — % is
6. Show that y = \/a; Jn(\x) is a solution of the equation
4&V + (4XV - 4n2 + I)?/ = 0.
7. Show that T/ = xnJn(x) is a solution of the equation
xy" + (I - 2n)y' + xy = 0.
8. Show that y = x~nJn(x) is a solution of the equation
xy" + (1 + 2n)i/7 + xy = 0.
101. Expansion in Series of Bessel Functions. It was pointed out
in connection with the development of arbitrary functions in trigo-
nometric series (Sec. 24) that the Fourier series development is only a
special case of the expansion of a suitably restricted class of functions
in series of orthogonal functions. It will be shown in this section that
it is possible to build up sets of orthogonal functions with the aid of
Bessel functions, so that one can represent an arbitrary function in a
series of Bessel functions.
It is shown in the treatises* on Bessel functions that the equation
Jn(x) = 0 has infinitely many positive roots Xi, \2, • • • , X», • • • ,
whose values can be calculated to any desired degree of accuracy. It
will be established next that the functions
\/xJn(\\x), -v/J/nCXaz), • • , \/xJn(\kx)t • • -
are orthogonal in the interval from x — 0 to x = 1, so that
(101-1) j^1 Vx J«(Xtz) • V* J*(\,x) dx = 0, if i 7* j,
= M<A/(Xt)]2, if i = j.
The proof of this fact depends on the following identity, the validity
of which, for the moment, will be taken for granted:
(101-2) (X2 - M2) xJn(\x)Jn(»x) dx
= x(iJ.
* See the references given in the footnote, p. 335.
340 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §101
Let X = X, and p, = Xt, where X» 7^ X,; then
J \/x Jn(\x) • \/z Jn(\jx) dx
Setting # = 1 and remembering that ./n(Xt) = Jn(X;) = 0 give the
first part of the formula (101-1).
In order to establish the second part, differentiate (101-2) partially
with respect to X, and thus obtain
(101-3) 2X xJn(\x)Jn(»x) dx + (X2 - M2) x\Jn(nx)Jn'(\x] dx
Set x = 1, X = /x, and recall that if X is a root of Jn(x) = 0 then
/n(X) = 0. Upon simplification of (101-3), one obtains the second pait
of the formula (101-1).
In order to establish the identity (101-2), note that y = \/xJn(\x)
is a solution of the equation*
4x*y" + (4\V - 4n2 + l)y = 0.
Setting u — \/x /n(Xx) and v = -\/x Jn(^x) gives the identities
+ (4X2^2 - 4n2 + l)u = 0
and
4x*v" + (4/z2^2 - 4n2 + 1> = 0.
Multiplying the first of these by v and the second by u and subtracting
furnish the identity
-(X2 - fj.2)uv = u"v - v"u.
The integration of both sides of this identity, between the limits 0 and
x, yields
- (X2 - /x2) f* uv dx = §* (u"v - v"u) dx
- [u\ - fo u'vt dx\ - [Ho - K uV
= [u'v - v'u]*Q.
Recalling the definitions of u and v gives the desired identity (101-2).
Since the formula (101-1) is established, it is easy to see that if f(x)
* See Prob. 6, Sec. 100.
§101 ORDINARY DIFFERENTIAL EQUATIONS 341
has an expansion of the form
/(*) =
which can be integrated term by term, then*
2 f1 xf(x)Jn(\tX) dx
The most common use of this formula is in connection with the expan-
sion of functions in a series of Bessel functions of order zero.
PROBLEMS
1. Show that -j- [xJ\(x)\ — xJQ(x).
2. Show that f(x) = 1, 0 < x < 1, when expanded in a series of
Bessel functions of order zero, gives
Hint: Make use of the results of Prob. 1 above and Prob. 1, Sec. 100.
3. Show that
—
dx [Xn< n(X)\ ~ Xn, n-lW
and
d
dx ' n*
4. Show that
[2 /sin x \
= V^ v"^ ~ cos v
and
6. Show, with the aid of the formulas of Prob. 3, that
and
* See Sec. 24.
342 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §102
*(h--}
6. Expand e2v */ in a power series in h to obtain
x(h l\ n^?
e^h I) = 2) A>s
n so — 00
and show that An = /»(#), so that
£/A_l\
4. Wl(j.) + to/l(a.) + . . . + frs/^) + . . . ]
+ [h-V-i(x) + /r2J_2(z) + • • • + h-»J-n(x) +•••].
102. Legendre's Equation.* The equation
(102-1) (1 - *2) g - 2* g + n(n + l)y = 0,
where n is a constant, occurs frequently in practical investigations
when spherical coordinates are used. One of the many uses of
this equation in practical problems is indicated in the next
chapter in connection with a study of the distribution of temper-
ature in a conducting sphere.
Assume, as in Sec. 100, that
(102-2) y = a0xm + aixm+l + • • • + akxm+l + • • •
is a solution of (102-1). Then,
d2v
-~5 = m(m — I)a0xm~2 + (m
+ (m + 2)(m
+ (m + k)(m + k -
d2y
-r~ = —m(m — I)a0xm — • • •
> - (m + k - 2)(m + k - 3)aA;_2xw+fc~2 -
~- -Q - • • • — 2(m + A? -
ax
n(n + l)i/ = n(n + l)aQxm + • • • + n(n
Adding these expressions and equating to zero the coefficients
of a;™"2, xm~l, • - - , xm+A;-2 give the system of equations
* A. M. Legendre (1752-1833), an outstanding French analyst who made
many brilliant contributions to the theory of elliptic functions. , /
§102 ORDINARY DIFFERENTIAL EQUATIONS 343
m(m — I)a0 = 0,
m(m + l)ai = 0,
(m + fc)(m + fc —
+ (n — m — k + 2)(n + m + k — l)ajb_2 = 0.
In order to satisfy the first of these equations, m can be chosen as
either 0 or 1. If m = 1, the second equation requires that
ai 55 0. For m = 0 the coefficient of zm+*~2 gives the recursion
formula
(n - k + 2)(n + fc - 1)
<** = -- k(k=-l) - a*-2>
from which, in a manner analogous to that employed in Sec. 100,
the coefficients a2, a3, a4, • • • can be determined. If w = 0,
the second of the equations of the system allows ai to be arbitrary.
If the values of the coefficients in terms of a0 and ai are sub-
stituted in (102-2), the following solution is obtained:
(102-3) „ - a. l - * + ~ 1)(n + 3)
3!
J- <n " l^U - 3^U + 2)^ + 4) .5 _
~l" 5!
It is readily shown by means of the ratio test that for non-
integral values of n the interval of convergence of the series in
(102-3) is ( — 1, 1). Moreover, since the first series in (102-3)
represents an even function and the second series represents an
odd function, the two solutions are linearly independent. The
sum of the two series, each multiplied by an arbitrary constant,
gives the general solution of (102-1), which is certainly valid if
|x( < 1. It can be verified directly that the choice of m = 1 does
not lead to a new solution but merely reproduces the solution
(102-3) with a0 = 0.
An important and interesting case arises when n is a positive
integer. It is clear that, when n is an even integer, the first series
in (102-3) terminates and reduces to a polynomial, whereas, when
n is an odd integer, the second series becomes a polynomial. If
the arbitrary constants a0 and a\ are so adjusted as to give these
polynomials the value unity when x = 1, then the following set
344 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §102
of polynomials is obtained:
Po(x) = 1,
Pi(x) = x,
/>,(*)- f *«-±,
P8(s) s - xs - ^ x,
r> / \ 7 • 5 A 0 5 • 3 9I3*1
2 * +'
4
9'7 7-5 5-3
\
where the subscripts on P indicate the value of n. Clearly, each
of these polynomials is a particular solution of the Legendre
equation in which n has the value of the subscript on P. These
polynomials are known as Legendre polynomials. They are
frequently used in applied mathematics. Very often, they are
denoted by Pn(cos 0), where cos 6 = x, so that, for example.
P3(cos 0) s jj cos3 B - ^ cos 0.
Zi Zi
The values of the Legendre polynomials (sometimes called surface
zonal harmonics) are tabulated* for various values of x.
A solution that is valid for all values of x outside the interval (— ,1, 1)
can be obtained by assuming it to have the form of a series of descending
powers of x. A procedure analogous to that outlined above leads to
the general solution of the form
n(n - 1) 0 , n(n - l)(n - 2)(n - 3)
r« __ - - — rn-2 4- — - - — - — - - r«~4
x 2(2n-l)x + 2 • 4(2n - l)(2n - 3) x
n(n - l)(n - 2)(n - 3)(n - 4)(n - 5) n_ n
2 • 4 • 6(2n - l)(2n - 3)(2n - 5) x ^ ' ' ' J
(n + l)(n + 2)
- ! — - ! ' --
2-4(2n-h3)(2n + 5)
(n + l)(n + 2)(n + 3)(n + 4)(n + 5)(n + 6)
2 • 4 • 6(2n + 3)(2n + 5)(2n + 7) - • • •
which is valid for |x| > 1, and where n is a positive integer.
* See JAHNKE, E., und F. EMDE, Funktionentaf eln ; BYERLY, W. E.,
Fourier Series and Spherical Harmonics.
§102 ORDINARY DIFFERENTIAL EQUATIONS 345
It will be shown next that the Legendre polynomials are orthogonal
in the interval from —1 to 1, so that they can be used to represent a
suitably restricted arbitrary function defined in the interval ( — 1, 1).
Note that (102-1) can be written in an equivalent form as
3j[(k-*Vl + »(» + i)tf-o,
and let Pm(x) and Pn(x) be two Legendre polynomials. Then,
[(1 - x*}Pm'(x)} + m(m + l)Pm(x) s 0
and
[(1 - x*)Pn'(x)\ + n(n + l)Pn(x) EE 0.
Multiplying the first of these identities by Pn(x) and the second by
Pm(x) and subtracting give
Pm(x) ^ [(1 - x*)P»'(x)] - Pn(x) ~ [(1 - x*)Pm'(x)\
+ (n-m)(n + m + l)Pm(x)Pn(x) = 0.
Integrating both members of this expression with respect to x between
the limits —1 and +1 gives the formula
Pm(x} ix [(1 ~ *2)p»'<*>] dx ~ S-i Pn(x} Tx l(l ~ X^P"'W **
+ (n - m)(n + m + 1) P Pm(x)Pn(x) dx = 0.
J ~ i
The application of the formula for integration by parts to the first two
integrals reduces this formula to
Pm(x)Pn(x) dx = 0.
Therefore,
J^ Pm(x)Pn(x) dx = 0, if m ^ n,
so that the Legendre polynomials are orthogonal.
It can be shown that
2m + 1
The derivation of this formula is somewhat tedious and will not be
given here.*
*See WHITTAKER, E. J., and G. N. WATSON, Modern Analysis, p. 305;
MAcRoBERT, J. M., Spherical Harmonics; Byerly, W. E., Fourier Series
and Spherical Harmonics, p. 170.
346 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §103
Consider next a function f(x) that is defined in the interval ( — 1,1),
and assume that it can be represented by a series of Legendre
polynomials
(102-4) /(*) = J£ arfn(x)
n = 0
that can be integrated term by term. It follows immediately from
Sec. 24 that the coefficients in the series (102-4) are given by the
formula
an =
J1 f(x)pn(x) dx, (n = 0, 1, 2, • - - ).
PROBLEMS
1. Show that the coefficients of hn in the binomial expansion of
(1 - 2xh + A2)-1'* are the Pn(x).
2. Verify that
P«M - (* - 1)-
by computing Pn(x) for n — 0, 1, 2, 3.
3. Expand f(x) — 1 + x — x1 in a series of Legendre polynomials.
4. Show that
P2«Gr) =P2»(-aO
and
Pjn + lCx) = -P2n+l(-aO.
6. Show that
p rm-r iv x ' 3 ' 5 ' ' ' (2n " 1}
P2n(0) - (-1)- 2 • 4 • 6 • • • 2n
and
P2n+l(0) = 0.
6. Show, with the aid of the formula
(1 - 2xh + h*)-» = ^Pn(x) (see Prob. 1),
n = 0
that
Pn(l) = 1
and
P.(-l) = (-1)-.
103. Numerical Solution of DilGferential Equations. The
method of infinite series solution of ordinary differential equations
affords a powerful means of obtaining numerical approxima-
tions to the solutions of differential equations, but its useful-
ness is limited by the rapidity of convergence of the series.
§103 ORDINARY DIFFERENTIAL EQUATIONS 347
Many differential equations occurring in physical problems
cannot be solved with the aid of the methods discussed in this
chapter, and one is obliged to resort to numerical methods.
Only one of these methods, which was developed by the French
mathematician E. Picard, is outlined in this section.*
Consider the problem of finding that particular solution of the
equation of first order,
(103-1) fx = /(*, »),
which assumes the value yQ when x = XQ. If both members of
(103-1) are multiplied by dx and the result is integrated between
the limits XQ and x, one obtains
or
(103-2) y = 2/0 + fxf(x, y) dx.
JXo
This is an integral equation, for it contains the unknown function
y under the integral sign.
Since the desired integral curve passes through (#o,,2/o)> assume
as a first approximation to the solution of (103-2) that y, appear-
ing in the right-hand member of (103-2), has the value 2/0- Then,
the first approximation to the solution of (103-2) is
yi(x) = 2/0 + \f(x, 2/0) dx.
Jxo
Performing the indicated integration gives y\ as an explicit
function of x, and substituting y\(x) in the right-hand member
of (103-2) gives the second approximation, namely
dx.
The process can be repeated to obtain
I/ate) = 2/0 + Fflx, yz(x)] dx,
Jxo
and so on. It is clear that the nth approximation has the form
« Vn(x) = 2/o + C*f[x, yn-\(x)} dx.
Jx^
* For other methods see Bennett, Milne, and Bateman, Numerical Inte-
gration of Differential Equations, Bulletin of National Research Council,
1933.
348 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §103
The functions 2/1(2), yi(x), • • • , yn(x) all take on the value
?/o when x is set equal to XQ, and it may happen that the successive
approximations y\(x)j 2/2(2), • • • , yn(x) improve as n increases
indefinitely; that is,
lim yn(x) = y(x)y
n— » QO
where y(x) is the solution of Eq. (103-1). It may be remarked
that, in order to establish the convergence of the sequence of
approximating functions, it is sufficient to assume that f(x, y)
and df/dy are continuous in the neighborhood of the point (XQ, 2/0).
Despite the fact that these conditions are usually fulfilled in
physical problems, the convergence may be so slow as to make
the application of the method impracticable. The usefulness of
the method is likewise limited by the complexity of the approxi-
mating functions. In many instances, it may be necessary to
make use of numerical integration in order to evaluate the
resulting integrals. *
The method just outlined can be extended to equations of
higher order.
As an illustration of the application of the method to a specific prob-
lem, let it be 'required to find the integral curve of the equation
y' = 2x + y\
passing through (0, 1).
Then (103-2) becomes
(103-3) y = i+f* (2x + I/*) dx,
and substituting y = 1 in the integrand of (103-3) gives
y\ = 1 + f* (2x + 1) dx = I + x + x\
Then,
2/2 = 1 + f* [2x + (1 + x + z2)2] dx
= 1 + x + 2x* + x* + y^ +
and
f
[2x + (l+x + 2x* + xs + %x* + Hx*Y] dx
+ 2x* + %x* + %x* + %x* + ylQx* + i%5V
+ %o*8 + %o*9 + Ko*10 + HTS*"- %
Even though the integrations in this case are elementary, the process
of computing the next approximation is quite tedious. As a matter of
* See Sec. 167.
J103 ORDINARY DIFFERENTIAL EQUATIONS 349
fact, in this case one can obtain the desired solution more easily by the
method of infinite series.
Thus, assuming
00
<«r\
anx*
and applying the method of Sec. 98 lead to the solution in the form
I 9 \ I 1 I 1 \ 9 I *-/t*u I ****U _
y = O,Q -f" #o x *T \^o i L)X* i ^ x -7-
o
5a03
-h
Since the integral curve must pass through (0, 1), it follows that a0 = 1,
and the desired solution is
y = 1 + x + 2z2 + y^ + iy*x* + ijhsz6 + • • ' .
This agrees with the solution obtained by Picard's method up to the
terms in x4.
PROBLEM
Find, by Picard's method, solutions of the following equations:
(a) y' = zi/ through (1,1);
(b) y' - x~y* through (0, K);
(c) y' = l+2/2 through g, l);
(d) y' = x + y through (1, 1).
CHAPTER VIII
PARTIAL DIFFERENTIAL EQUATIONS
104. Preliminary Remarks. An equation containing partial
derivatives has been defined in Sec. 67 as a partial differential
equation. This chapter contains a brief introduction to the
solution of some of the simpler types of linear partial differential
equations which occur frequently in practice. It will be seen that
the problem of solving partial differential equations is inherently
more difficult than that of solving the ordinary equations and
that Fourier series, Bessel functions, and Legendre polynomials
play an important part in the solution of some of the practical
problems involving partial differential equations.
It was stated in Sec. 68 that the elimination of n arbitrary
constants from a primitive /(x, y, Ci, c2, • • • , cn) leads, in
general, to an ordinary differential equation of order n. Con-
versely, the general solution of an ordinary differential equation
has been defined to be that solution which contains n arbitrary
constants. In the next section, it is indicated, by some examples,
that one is led to partial differential equations by differentiating
primitives involving arbitrary functions, and it follows that
partial differential equations may have solutions which contain
arbitrary functions. However, it is not always possible to
eliminate n arbitrary functions from a given primitive by n
successive differentiations, and the temptation to define the
general solution of a partial differential equation as the one con-
taining n arbitrary functions may lead to serious difficulties.
In some important cases of linear partial differential equations
with constant coefficients, treated in Sec. 107, it is possible to
obtain solutions that contain the number of arbitrary functions
equal to the order of the differential equation, and the term
general solution is used in this chapter only in connection witfy
such equations. With the exception of the linear partial differen-
tial equations of the first order and of certain important types of
linear equations of the second order, no extensive theory of the
nature of solutions has been developed so far.
350
§105 PARTIAL DIFFERENTIAL EQUATIONS 351
Just as in the case of the ordinary differential equations, the
solution of a practical problem can be obtained by eliminating
the element of arbitrariness with the aid of the initial or boundary
conditions. In practical problems the boundary conditions
frequently serve as a guide in choosing a particular solution,
which satisfies the differential equation and the boundary con-
ditions as well.
105. Elimination of Arbitrary Functions. Consider a family of
surfaces defined by
*=/(* + y),
where / is an arbitrary function.
If the argument of / is denoted by s, then
*=f(x + y) -/(«)
and
dz _ df ds
dx ~~ ds dx
Since s = x + y, it follows that
(105-1) g = f(x + y},
where f'(x + y) denotes the derivative of f(x + y) with respect to
its argument x + y. Similarly,
(105-2) g==/>(x + 2/).
Subtracting (105-2) from (105-1) leads to the partial differential
equation of the first order
^_*£ = o
dx dy '
whose solution, clearly, is z = f(x + y).
If
-'GO
then
dx ds dx dy ds dy
where s = y/x.
352 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §106
Denoting df/ds by f(y/x) and substituting the values of
ds/dx and ds/dy give
- f> - and -
- ' *2/ ~
from which /'(y/x) can be eliminated to give
dz . dz A
z — + 2/ — = 0.
do; d?/
Again the result is a partial differential equation of the first order.
On the other hand, if
*=/i(*)+/2(y),
where f\(x) and fa(y) are arbitrary functions, differentiations
with respect to x and y give
!=/((*) and !=/;(,).
If the first of these relations is differentiated with respect to y,
a partial differential equation of the second order results, namely,
dx dy
=0.
Differentiation of the second relation with respect to x will lead
to the same equation, for the derivatives involved are assumed
to be continuous.
Another example, which is of considerable importance in the
theory of vibrations, will be given. Let
z =fi(x + at) +f2(x - at).
If x + at ss r and x — at = s, then
and
,
OX uT uX uS uX
"== J 1\*^ "i ^'/ "T~ J 2\*^ "~~ Qtt) •
Similarly,
(105-3) d~ = f!(z + at) + ft(x - at).
§106 PARTIAL DIFFERENTIAL EQUATIONS 353
Also,
dz a(/i +/2) dr a(
dZ 3r dt ds dt
= fi(x + a()a +ft(x - at)(-a)
and
(105-4) g = fi(x + a«)a» + K(x - a*)a*.
Eliminating /'/(z + at) and/^'(a: - aO from (105-3) and (105-4)
gives
&z _ 2M
dt* " a dx*'
*
regardless of the character of /i and /2. This partial differential
equation is of primary importance in the study of vibration and
will be considered in more detail in Sees. 106, 108, and 109.
106. Integration of Partial Differential Equations. This sec-
tion contains two examples illustrating integration of partial
differential equations.
Let the differential equation be
Integration with respect to y gives
(106-2) ^ = /(*),
where f(x) is arbitrary. If (106-2) is integrated with respect to
x, then
where ^ and <p are arbitrary functions.
Consider next
nnr *\ &z *&*
(106-3) a?~a 5?
Change the variables in this equation by setting r = x + at
and s = x — at so that z becomes a function of r and s. Then
dz = dzdr , ^^£.
dx "" dr dx ds dx'
354 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §106
and since dr/dx « 1 and ds/dx = 1, it follows that
d£ _ dz dz
dx ~~ dr "*" ds
and
d2z /dzzdr , d2z ds , d2z ds
Similarly,
and since dr/dt = a and ds/d£ = — a, it follows that
dz _ dz __ dz
Differentiating this with respect to t gives
/-™ -\ d2z d2z dr , d2z ds d2z ds d2z dr
(106-5) — 2 == a — 2 1" a a — 2 a —
Substituting (106-4) and (106-5) in (106-3) gives the equation
dTds = °'
which is of the type (106-1), whose solution was found to be
z = \[/(r) + <p(s), where ^ and <p are arbitrary. Recalling that
r = x + at and s = x — at, it is seen that
(106-6) z = $(x + at) + <p(x - at),
which is the solution in terms of the original variables. If in
this solution \l/ and <p are so chosen that
\l/(x + at) s A sin k(x + at),
<p(x — at) s A sin k(x — at),
where the variable t is thought to represent the time and x is the
distance along the #-axis, then the first of these equations repre-
sents a sinusoidal wave of amplitude A and wave length X = 2ir/k
which is moving to the left with velocity a, whereas the second
§106 PARTIAL DIFFERENTIAL EQUATIONS . 355
expression represents a similar wave moving with velocity a to
the right (see Fig. 92). This can best be seen by recalling that the
replacement of x by x — at in z = f(x) shifts the curve at units
in the positive direction of x and that the substitution of x + at
for x translates the curve z = f(x) at units in the opposite direc-
z*A$inkx
FIG. 92.
tion. Since t is a continuous variable representing the time,
it is clear that the expression
A sin k(x — • at)
states that the sinusoid
z = A sin kx
is advancing in the positive direction of the x-axis with the
velocity a. The period of the wave
z = A sin k(x — at)
is defined as the time required for the wave to progress a distance
equal to one wave length, so that
X = aT
or
T = - = — -
a ka
Consider next the wave resulting from the superposition of the
two moving sinusoids A sin k(x — at) and A sin k(x + of).
Then,
z = A sin k(x — at) + A sin k(x + at)
= A (sin kx cos kat — cos kx sin
+ 4 (sin kx cos fca£ + cos kx sin
or
(106-7) z = (2 A cos kat) sin to.
The expression (106-7) is frequently referred to as a standing
wave, because it may be thought of as representing a sinusoid
356 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §106
sin kx whose amplitude 2A cos kat varies with the time t in a
simply harmonic manner. Several curves
z = 2 A cos kat sin kx
are drawn in Fig. 93 for various values of t. The points
x = '
= o, i, 2, • • • ),
are stationary points of the curve and are called nodes.
Inasmuch as (106-7) is obtained from (106-6) by making a
particular choice of \f/ and <?, it satisfies the differential equation
xZ«2Acoskoit
Fia. 93.
(106-3), whatever be the values of A and k. This fact is of great
importance in the discussion of Sec. 108.
PROBLEMS
1. Form partial differential equations by eliminating arbitrary
functions.
(a) z=f(x- 2y)
(6) *=/(*» + y»
3x + 4y.
Note that ~ = f'(x2 + ?/ + z2) (2x + 2z —
ox \ ox
(c) z
(d) z = /1(x)/2(?/).
2. Prove that z - fi(x + iy) + f%(x — iy) is a solution of
d^"2 + dy* = °'
3. Form partial differential equations by eliminating the arbitrary
functions, in which x and t are the independent variables.
(a) z =/!(* -20 +/>(* + 20;
(6) ««/(»-« + «);
(c) *-/i(* + 20 +/,(* + W);
§107 PARTIAL DIFFERENTIAL EQUATIONS 357
(d) z »# 1(3 +
(/) 2 = /i(s - 0 + */2(* - 0-
4. Show that z = f(a\y — a«x) is a solution of the equation
dz dz
aidx + a*frj = ">
where a\ and a2 are constants.
5. Verify that z = j\(y + 2x) and z = f»(y — 3ar) satisfy the equation
**+_*'*__ 6*!i -o
ax2 + dxdy a?/2 ~ u'
and hence deduce that z = fi(y + 2«) + /2(?/ — 3x) is also a solution of
the equation.
6. Show that
is a solution of the equation
2±, 2 _*?!..££ -o
ax4 "*" ax2 a^/2 "^ a?j4 u>
provided that i2 = — 1.
107. Linear Partial Differential Equations with Constant
Coefficients. A linear partial differential equation with constant
coefficients that often occurs in applications has the form
(107-1) ^o + al + a22
Frequently, such equations are called " homogeneous " because
they involve only derivatives of the nth order.
This equation can be solved by a method similar to that
employed in solving an ordinary linear equation with constant
coefficients. Introduce the operators
DJ - £ and D, - £,
with the aid of which (107-1) can be written as
(107-2)
358 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §107
It is readily established* that the operators D\ and D2 formally
obey the ordinary laws of algebra, so that one can deal with
differential operators of the form
D2) s
just as one would with polynomials in the two variables DI and
D2. Accordingly, the left-hand member of (107-2) can be
decomposed into a product of n linear factors, so that (107-2)
reads
(107-3) (aiDi + 0iD*)(aJ)i + jW>,) • - - (a»Di + /U>2)z = 0,
where the quantities a, and ft, in general, are complex numbers.
Now the system of equations
(alDl + ftZ>2)z = 0, (z = 1, 2, • • • , n),
or
(107-4) «,|1 + A|| = o, (i = 1, 2, • - • , n),
can be readily solved. It is easy to verify that
z = Ft(aty - ftz),
where /*\ is an arbitrary function, is a solution of (107-4). Con-
sequently,! the solution of (107-3) can be written in the form
(107-5) z =
If the linear factors appearing in (107-3) are all distinct, the
solution (107-5) contains n arbitrary functions and will be called
the general solution of (107-1).
If the at in (107-3) are all different from zero, one can write
(107-3) as
(107-6) (Di - miJD,)CDi - m2D2) • • • (Di - mj)2)z = 0,
where ra» = —ft/at, (i = 1, 2, • • • , n). In this case, (107-5)
assumes the form
(107-7) z = Ft(y + mlx) + F2(y + mzx) + • • •
+ Fn(y + mnx).
If some of the factors in (107-6) are alike, then the number of
arbitrary functions in (107-7) will be less than n, but it is easy
* See the corresponding discussion in Sec. 87.
t See the corresponding case in Sec. 88.
J107 PARTIAL DIFFERENTIAL EQUATIONS 359
to see that the equation
(Di - wZ>2)rz = 0
has the solution
z = Fi(y + mx) + zF2(i/ + mx) +•-,-+ xr~lFr(y + mx).
Consequently, one can obtain a solution of (107-6) that contains
the number of arbitrary functions equal to the order of the
differential equation even in the case when some of the factors
in the left-hand member of (107-6) are not distinct.
As an illustration, consider the equation, which frequently
occurs in the study of elastic plates,
* * d*z
* '
dx* dx* dy2 dy*
or
(Z>i + 2D\D\ + £»|)z = 0.
The decomposition into linear factors gives
(Di + iD2)(D1 - iD*)(Di + iD*)(Di - */>»)« = 0,
where z2 = — 1. It follows that the general solution of this
equation has the form
z = Ft(y - ix) + xF*(y- ix) + F3(y + ix) + xF*(y + ix).
If the right-hand member of (107-1) is a function /(x, t/), then
the general solution of the equation is
z = $(z, y) + u(x, y),
where u(x, y) is a particular integral and $(#, y) is the general
solution of the related homogeneous equation. The determina-
tion of particular integrals of the equation
(107-8) L(Dl9 D2)z = f(x, y)
can be made to depend on the calculus of operators* as was done
in Sec. 89. In many cases the particular integral can be obtained
by inspection. If f(x, y) is a homogeneous polynomial of degree
k, then the particular integral has the form
(107-9) z = c0xk+n + dxk+n-ly + • • - + ck+nyk+n,
in which the coefficients cl can be determined by substituting
(107-9) into (107-8) and comparing the coefficients of the corre-
* See, for example, M. Morris and O. Brown, Differential Equations, p.
294; A. Cohen, Differential Equations, p. 275.
360 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §107
spending terms of the resulting equation.
As an example of this, consider
which can be written as
or
(/>! - 2Z)2)(Z>i + 3D2)z
Assume the particular integral in the form
(107-11) z = coz5 + cix4y + w*y* + c*x'2y3 + cay4 +
Substituting (107-11) in (107-10) gives
(20c0 + 4ci - 12c2)x8 + (12ci + 6c2 - 3(ic
_i_ ffj^o [ f\(* , — 72c4^x?y'2 •••( (2ci I 4c^i 120f>R>)?/^ ~— fJx'^/y
Hence, equating the coefficients of like terms on both sides of this
equation gives
5c0 -f- ct — 3c2 = 0,
12ci + 6c2 - 36cs = 6,
c2 + c3 — 12c4 = 0,
c3 + 2c4 - 60c5 = 0.
This system of four equations in six unknowns can always be solved.
Writing it as
Ci — 3c2 -f- Oc3 + Oc4 = — 5c0,
2ci + c2 •— 6c3 + Oc4 = 1,
Oci + c2 + c3 — 12c4 = 0,
Oci + Oc2 + c3 + 2c4 = 60c5,
and solving for ci, ca, c3, and c4 in terms of Co and c5 give a two-parameter
family of solutions,
-65c0 + 6480c6 + 21
C2
C3 =
C4 =
55
70c0 + 2160c5 + 7
55 '
-10co + 2520c6 -
55
78Qc5 + IQco + 1
110
Setting c0 = Cs = 0 and substituting the values of the coefficients in
(107-11) give a particular integral of (107-10) in the form
§108 PARTIAL DIFFERENTIAL EQUATIONS 361
«(*, y) = *K*x*y + %5*V - x**V
Therefore the general solution of (107-10) is
* = Fl(y + 2x) + F2(y - 3s) + u(x, y),
where FI and F2 are arbitrary functions.
PROBLEMS
1. Find the general solutions of
2. Find the particular integrals for the following equations:
r \ o <£* . ^ ^ _ i
W ^ Sx2 + dx dy ~ dy* ~ 1;
(b)**»z_Vz
^ ' dx2 dxdy dy2 " '
//zn^: Obtain the particular integral for /(a;, y) = y2 and for/Or, ?/) = a:
and add the solutions.
to ^ + 3 A22. , 2— -^4-77-
(c) dx2 ^*dxdy^ Zdy2 X^y'
(d]^_a^-x*
W dx2 a dy2 ~ x '
108. Transverse Vibration of Elastic String. Consider an
elastic string or wire of length / stretched between two points on
the x-axis that are I units apart, and let it be distorted into some
curve whose equation is y — f(x) (Fig. 94). At a certain instant,
say t = 0, the string is released from rest and allowed to vibrate.
The problem is to determine the position of any point P of the
string at any later time t. It is assumed that the string is per-
fectly elastic and that it does not offer any resistance to bending.
The resulting vibration may be thought of as being composed of
the two vibrations:
362 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §108
a. Transverse vibration, in which every particle of the string
moves in the direction of the t/-axis;
6. Longitudinal vibration, in which every particle oscillates
in the direction of the rr-axis.
It is tolerably clear that, if the stretching force T is large com-
pared with the force of gravity, then the horizontal component
of tension in the string will be sensibly constant. Therefore the
displacement of the point P in the direction of the tf-axis can be
neglected compared with the displacement of P in the y-direction.
In other words, the longitudinal oscillation of the string con-
tributes so little to the resultant vibration that the entire vibra-
tion may be thought to be given by considering the transverse
component-vibration.
•T+AT
FIG. 94.
The relation connecting the coordinates of the point P with
the time t can best be stated in the form of a differential equation.
Thus, denote the length of the segment of the string between the
points P(XJ y) and Q(x + A£, y + AT/) by As, and let the tension
at P be T and at Q be T + A7T. In view of the assumption
stated above, the horizontal components of tension at P and Q
are nearly equal so that the difference AT of the tensions at the
ends of the segment As is taken as equal to the difference between
the vertical components of tension at Q and P. The vertical
component of tension at P is
(A \ / ^ \
T lim =) = (T^J ,
As_oAs/p \ ds/x
and the vertical component of tension at Q is
(V
TIT)
OS/x+bx
If it is assumed that the transverse displacement of the string
is so small that one can neglect the square of the slope of the
string in comparison with the slope dy/dx, then the sine of the
§108 PARTIAL DIFFERENTIAL EQUATIONS 363
angle can be replaced by the tangent,* and the resultant of
the forces at P and Q is
By Newton's second law, this resultant must equal the mass of
the element of the string of length Ax multiplied by the accelera-
tion in the direction of the ?/-axis. Hence,
<"»> "*(&),-«• [(£
where p is the mass per unit length of the string and
denotes the acceleration of the element PQ of the string.
Dividing both sides of (108-1) by p &x reduces it to
"
and passing to the limit as Ax— »0 gives
where a2 = T/p.
The solution of (108-2) was found in Sec. 106 to be
y = $(x + at) + v(x - at),
where ^ and <p are arbitrary functions. These functions must
be so chosen that, when t = 0,
Represents the equation of the curve into which the string was
initially distorted. Furthermore, the string was supposed to
have been released from rest, so that dy/dt = 0 when t = 0. It is
beyond the scope of this book to prove that these boundary
conditions suffice for the unique determination of the functions
* Note that
dy
dy . „ tan 6 dx ^ dy
~ « sin 0 "~ '
VI + tan2 0
364 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §109
\// and <p. It will be shown in the next section how the solution
of this problem is obtained with the aid of Fourier series.
109. Fourier Series Solution. In the preceding section, it was
established that the transverse vibrations of an elastic string are
defined by the equation
and in Sec. 106 it was shown that a particular solution of this
equation is given by
(109-2) y = 2A cos kat sin kx
for arbitrary values of A and k. Moreover, it is clear that the
sum of any number of solutions of the type (109-2) will satisfy
(109-1).
Now, suppose that the string of length I is distorted into some
curve y = /Or), and then released without receiving any initial
velocity. The subsequent behavior of the string is given by
Eq. (109-1), the solution of which must be chosen so that it
reduces to y = f(x) when t = 0. In addition to this condition,
dy/dt — 0 when t = 0, for, by hypothesis, the string is released
without having any initial velocity imparted to it. Furthermore,
since the string is fixed at the ends, y = 0 when x = 0 and when
x = l.
Consider the infinite series
/lrkrk ON vat . irx . 2irat . 2irx
(109-3) y = di cos -y- sm -j- + #2 cos — j— sm — =-
II 11
3wat . 3wx ,
+ a3 cos -y— sm -j -- 1- • • • ,
each term of which is of the type (109-2), where k has been chosen
so that each term reduces to zero when x = 0 and when x = I.
When t = 0, the series becomes
/^™ A\ . wx . . 2irx . . .
(109-4) ai sm y + a2 sin ~~T — I" as sin —j — f-
If the coefficients an are chosen properly, (109-4) can be made to
represent the equation y = f(x) of the curve into which the
string was initially distorted; for a function /(#), subject to
certain restrictions,* can be expanded in a series of sines (109-4)
* See Sec. 20.
§109
PARTIAL DIFFERENTIAL EQUATIONS
365
and the coefficients are given by
(109-5) «n = r f /(*) sin ^r dx-
v ' I Jo J I
It is readily verified that the derivative of (109-3) with respect
to t satisfies the remaining boundary condition, dy/dt = 0
when t = 0. Hence the infi-
nite series (109-3), where the
values of an are given by
(109-5), gives the formal solu-
tion of the problem.
Illustration. If the initial dis-
tortion of the string (Fig. 95) is
given by
26
k
FIG. 95.
then the solution of the problem is readily found to be
86/1 . TTX
irat
~ ~~2 V ?2 sm "T cos ~T~
7T \ 1 I I
1 .
— 7™ Sin
—j— cos — j— +
i I
PROBLEMS
1. Carry out solution of the problem given in the illustration, Sec. 109.
2. A taut string of length /, fastened at both ends, is disturbed from
its position of equilibrium by imparting to each of its points an initial
velocity of magnitude f(x) . Show that the solution of the problem is
y = sm
sm
sm
Hint: The schedule of conditions here is:
(a) y = 0, when t = 0;
(6) -| = /(*), when « = 0;
(c) T/ = 0, when x = 0;
(rf) ?/ = 0, when a; = /.
Observe that
, . .
A sm — T- sm
nirat
366 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §109
satisfies conditions (a), (c), and (d), and build up a solution by forming
mrx . mrat
SA . nirx .
An sm — T- sir
and utilizing condition (6).
3. Show that the solution of the equation of a vibrating string of
length /, satisfying the initial conditions
y — f(x)> when t = 0, and r = g(x), when t = 0,
niral , ^\ . . mrx . nwal
cos —7 h X on sm —j- sin — T— >
•, v li ^^i ' '
n-1 n=l
is
y = 2) «„ sin -y-
where
= - r i -
and
2
— I /t^'y»^ oif» .^
/
4. The differential equation of a vibrating string that is viscously
damped is
Show that the solution of this equation, when the initial velocity is
zero, has the form
y =
where
ane~bt sin -7- ( cos anl + — sin anl) ,
L * \ «n /J
70 . 2 rl „ ^ - mrx .
~" ^ anc^ a» ~ 7 I f(x) sm ~7~ ^a'-
6. Show that the differential equation of the transverse vibrations of
an elastic rod carrying a load of p(x) Ib. per unit length is
d±y __ p(x) <Py
dx* " El m dl* '
where E is the modulus of elasticity, / is the moment of inertia of the
cross-sectional area of the rod about a horizontal transverse axis through
the center of gravity, and m is the mass per unit length.
Hint: For small deflections the bending moment M about a horizontal
transverse axis at a distance x from the end of the rod is given by the
Euler formula M == El d2y/dx*, and the shearing load p(x) is given
by d*M/dx* = p(x).
§110 PARTIAL DIFFERENTIAL EQUATIONS 367
6. Show that the small longitudinal vibrations of a long rod satisfy
, the differential equation
_
dt* ~ p dx2'
where u is the displacement of a point originally at a distance x from
the end of the rod, E is the modulus of elasticity, and p is the density.
Hint: From the definition of Young's modulus E, the force on a cross-
sectional area q at a distance x units from the end of the rod is
Eq(du/dx)x, for du/dx is the extension per unit length. On the other
hand, the force on an element of the rod of length Ax is pq Ax d2u/dt2.
1. If the rod of Prob. 6 is made of steel for which E = 22 • 108 g.
per square centimeter and whose specific gravity is 7.8, show that the
velocity of propagation of sound in steel is nearly 5.3 • 105 cm. per
second, which is about 16 times as great as the velocity of sound in air.
Note that the c.g.s. system E must be expressed in dynes per square
centimeter.
110. Heat Conduction. Consider the slab cut from a body
r by two parallel planes As units apart, and suppose that the
temperature of one of the planes is u and that of the second plane
is u + At£. It is known from the results of experiments that heat
will flow from the plane at the higher temperature to that at the
lower and that the amount of heat flowing across the slab, per
unit area of the plane per second, is approximately given by
(110-1) kg,
where k is a constant called the thermal conductivity* of the
substance. If the distance As between the planes is decreased,
then the limit of (110-1),
,. 7 Aw 7 du
hm k — = A; — >
AS->O As as
gives the quantity of heat flowing per second per unit area of the
surface whose normal is directed along s, and the quantity
du/ds gives the rate of change of temperature in the direction of
increasing s.
Now suppose that the initial temperature of such a body is
given by
and that it is required to find the temperature of the body at
* The dimensions of k in the c.g.s. system are cal./(cm.-sec. °C.).
368 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §110
some later instant i. It is known* that the function u, which
gives the temperature at any later time t, must satisfy the partial
differential equation
(110-2) £_*
v ' dt Cp
where c is the specific heat of the substance, p is the density
of the body, and k is the conductivity, f Equation (110-2) is
derived on the assumption that k, c, and p remain independent
of the temperature u, whereas in reality they are not constant
but vary slowly with the temperature. Moreover, this equation
is not true if there is any heat generated within the body.J
This equation must be solved subject to certain boundary
conditions.
Thus, if the body is coated with some vsubstance which makes
it impervious to heat so that there is no heat flow across the sur-
face of the body, then, if the direction of the exterior normal to the
body is denoted by n, this boundary condition can be expressed
mathematically as
1 du A du n
k — = 0 or — = 0.
dn dn
On the other hand, if the surface of the body radiates heat
according to Newton's law of cooling, § then
i du ( \
*- = *(„-«,),
where u0 is the temperature of the surrounding medium and e is
a constant called the emissivity of the surface. It can be shown ||
that, if the initial and surface conditions are specified, then the
problem of determining the temperature at any later time t has
only one solution.
It should be observed that if the flow of heat is steady, so that
the temperature u is independent of the time t, then du/dt - 0
* See derivation of this equation in Sec. 130.
f The dimensions of c and p are, respectively, in calories per gram per
degree centigrade and grams per cubic centimeter. The constant k/cp — a2
sq. cm. per second is frequently called the diffusivity.
t See Sec. 130.
§ See Problem 2, Sec. 70.
|| For detailed treatment see Carslaw, "Introduction to the Mathematical
Theory of the Conduction of Heat in Solids."
§111
PARTIAL DIFFERENTIAL EQUATIONS
369
and (110-2) reduces to
(110-3)
dx2
^
dy* ^ dz*
This is known as Laplace's equation, and it occurs frequently
in a large variety of physical problems.
It may be remarked that the problems of diffusion and the
drying of porous solids are governed by an equation similar to
(110-2), so that many problems on diffusion and heat conduction
are mathematically indistinguishable.
111. Steady Heat Flow. Consider a large rectangular plate
of width d, one face of which is kept at tem-
perature u = Ui, whereas the other face is kept
at temperature u = u^. If one face of the
plate is placed so as to coincide with the
2/2-plane (Fig. 96), the surface conditions can
be expressed mathematically as
u«u,
(111-1)
n — HI when x = 0,
u = u<t when x — rf,
and the temperature u must satisfy Eq. -
(110-3). In this formulation of the problem,
it is assumed that the plate extends indefi-
nitely in the y- and ^-directions, a condition
that is approximated by the large rectangular
plate if the attention is restricted to the
middle of the plate. With these assumptions,
it is clear that the temperature u is inde-
pendent of the y- and ^-coordinates and that (110-3) reduces to
U-Ug
FIG. 96.
(111-2)
dx*
= 0,
which is to be solved subject to the conditions (111-1).
The solution of (111-2) is easily found to be
(111-3) u = cix + c2,
where Ci and c2 arc arbitrary constants which must be determined
so that (111-3) satisfies (111-1). Substituting x = 0 and x = d
in (111-3) gives u\ = c2 and it2 = c\d + c2, so that
u =
-x +
370 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §111
gives the solution of the problem. Recalling that the amount
of heat flowing per second per unit area of the plate is
7 du i Uz — Ui
k — = k 1 — i
dx d
it is seen that the amount of heat flowing in t sec. over the area
y A is given by
_ i.
- u^tA.
«-i
These results can be anticipated from physical
considerations.
A more difficult problem will be solved next.
Suppose that a " semi-infinite " rectangular plate
(that is, the plate extends indefinitely in the
* positive direction o the ?/-axis), of thickness d,
has its faces kept at the constant temperature
FIG. 97. u = 0, whereas its base y — Q is kept at tem-
peratujre u = f(x) (Fig. 97). It is clear physically that the tem-
perature u at any point of the plate will be independent of 2, so
that in this case (110-3) becomes
u»f(x)
(111-4)
dx* + dy*
(111-5)
The solution of (111-4) must be so chosen that it satisfies the
boundary conditions:
u = 0 when x = 0,
u — 0 when x = d,
u = f(x) when y = 0,
u = 0 when y = oo.
The last condition results from the observation that the tem-
perature decreases as the point is chosen farther and farther from
the a>axis.
In order to solve (111-4), recourse is had to a scheme that
often succeeds in physical problems. Assume that it is possible
to express the solution of (111-4) as the product of two functions,
one of which is a function of x alone and the other a function of y
alone. Then,
(111-6) u = X(x)Y(y).
Substitution of (111-6) in (111-4) and simplification give
§111 PARTIAL DIFFERENTIAL EQUATIONS 371
= _.
j X dx2 Y dy2
It will be observed that the left member of (111-7) is a function
of x alone, whereas the right member is a function of y alone.
Since x and y are independent variables, Eq. (111-7) can be true
only if each member is equal to some constant, say — a2. Hence,
(111-7) can be -written as
1 d2X 2 ,1 d2Y 2
•v~rr = —a and 17 -FT = a
X dx2 Y dy2
or
~ + a2X = 0 and a2Y = 0.
dx2 dy2
The linearly independent solutions* of these equations are
X — sin ax,
X — cos ax,
Y = eav,
Y = e-a",
and, since u = XY, the possible choices for u are
u =
eoj/ cos ax,
eay sin ao:,
The first two of these particular solutions for u obviously do not
satisfy the last one of the boundary conditions (111-5). The
third particular solution e~ay cos ax does not satisfy the first
of the conditions (111-5). But if u is chosen as e~ay sin ax, then
u = 0 when x = 0 and u = 0 when y = oo ; and if a is chosen
as nir/d, where n is an integer, then
/i 1 1 o\ ~^ry •
(111-8) u = e d sin
/
satisfies all the conditions (111-5), except u = /(#) when y = 0.
It will satisfy this condition also if f(x) = sin ~™
Inasmuch as Eq. (111-4) is linear, any constant times, ^ solu-
tion (111-8) will be a solution, and the sum of any number ,pf such
* See Sec. 95.
372 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §111
solutions will be a solution. Hence,
00 niry
(111-9) « = ^ ««e~^" sin ^
n=>l
is a formal solution. When y = 0, (111-9) becomes
00
2. nwx
an sin —T- >
n = l
which must reduce to/(x). But, in Sec. 20, it was shown that
the constants an can be chosen so that the function is represented
by a series of sines. Therefore, if
then (111-9) will satisfy all the boundary conditions of the prob-
lem and hence it is the required solution.
Illustration. In the preceding problem suppose that f(x) — 1 and
d = TT. Then
2 r* -
a» = - I sin nx ax,
TT JO '
and the solution (111-9) is easily found to be
4 / 1 1 \ •*
u = - ( e~tf sin x + ^ e~?J/ sin 3z + ^ e~5tf sin 5x + • • • J.
PROBLEMS
1. Using the result of the illustration just above, compute the tem-
peratures at the following points: (ir/2, 1), (?r/3, 2), (?r/4, 10).
2. Obtain the solution of the problem treated in Sec. 109 by assuming
that y can be expressed as the product of a function of x alone by a
function of t alone and following the arguments of Sec. 111.
3. Compute the loss of heat per day per square meter of a large con-
crete wall whose thickness is 25 cm., if one face is kept at 0°C. and the
other at 30°C. Use k = 0.002.
4. A refrigerator door is 10 cm. thick and has the outside dimensions
60 cm. X 100 cm. If the temperature inside the refrigerator is — 10°C.
and outside is 20°C. and if k = 0.0002, find the gain of heat per day
across the door by assuming the flow of heat to be of the same nature
as that across an infinite plate.
5. A semi-infinite plate 10 cm. in thickness has its faces kept at 0°C.
and its base kept at 100°C. What is the steady-state temperature at
any point of the plate?
§112 PARTIAL DIFFERENTIAL EQUATIONS 373
112. Variable Heat Flow. Consider a rod of small uniform
cross section and of length /. It will be assumed that the surface
of the rod is impervious to heat and that the ends of the rod are
kept at the constant temperature u = 0°C. At a certain time
t = 0, the distribution of tempera-
ture along the rod is given by
y = /(#)• The problem is to find the
temperature at any point x of the
rod at any later time t.
In this case the temperature u is
a function of the distance along the FlG> 98>
rod and the time £, so that, if the rod (Fig. 98) is placed so as to
coincide with the z-axis, (110-2) becomes
N du n d2u
where a2 = k/cp is the diffusivity. In addition to satisfying
(112-1) the solution u must satisfy the boundary conditions
!' u = 0 when x = Q) f „ , r .
A , i \ f or all values of t,
u = 0 when x = I j
u = f(x) when t = 0.
As in Sec. Ill, assume that a solution of (112-1) is given by the
product of two functions, one a function of x alone and the other
a function of t alone. Then,
u = X(x)T(f), .
and the substitution of this expression in (112-1) gives, after
simplification,
1 d_ T = 1 d*X
a*T dt ~ X ~dx* '
This equation can hold only if each member of it is equal to some
constant, say — /32. There result
+ a^T = 0 and + pX = 0.
at ax*
The linearly independent solutions of these ordinary differential
equations are readily found to be
X = cos fix,
X = sin ftx.
374 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §112
Then, since by hypothesis u = TX, the possible choices for u are
u = e~a^2t cos ftx,
u = e-«*w sjn pXm
The first particular solution does not satisfy the first one of the
conditions (112-2). If ft is chosen as mr/l, where n is an integer,
then
(112-3) u-e-'W'xin^x
satisfies the first two conditions of (112-2) but not the last one.
The sum of solutions of the type (112-3), each multiplied by a
constant, will be a solution of (112-1), since the equation is linear,
so that
(112-4) M = 2±ane V ' / sin ^-x
is a solution. For t = 0, (112-4) reduces to
oo
2. nit
an sin -y x,
which can be made equal to/(x), provided that
= ? fl /(j.) 8in !^5 da,
/ Jo £
Then,
(H2-5) u = 2 [r jo'/(x) sin ^ r/x j 6 ^ ' sin T
satisfies all the conditions of the problem and is therefore the
required formal solution.
Next, consider an infinite slab of thickness I, whose faces are
kept at temperature zero and whose temperature in the interior
at the time t = 0 is given by u = f(x). It is clear that the
solution of this problem is independent of y and z, so that u satisfies
the differential equation
— — 2 ^u
~di ~~ "" daT2"
The boundary and initial conditions are
u = 0 when x = I ' ™ '
u = f(x) when / = 0.
§112
PARTIAL DIFFERENTIAL EQUATIONS
375
The mathematical formulation of this problem is identical with
that of the preceding one, and therefore the solution of the
problem is given by (112-5).
The solutions of other important problems on heat flow are
outlined in detail in Probs. 5 and 6 at the end of this section.
PROBLEMS
1. Suppose that in the first problem of Sec. 112 the ends of the rod
are iniDervious to heat, instead of being kept at zero temperature.
The formulation of the problem in such a case is
du _ 23%
dt ~ a dx2'
du - n
~r~ — U
dx
when x — 0
- -0
dx ~ U
for all values of t,
when x — I
u = f(x) when t = 0.
Show that the solution in this case is '
00
u — --
where
2 fl
= j- J0
cos
cos
,
dx.
o°
100°
01
0°
Fia 99.
2. A large rectangular iron plate (Fig. 99) is
heated throughout to 100°C. and is placed in con-
tact with and between two like plates each at 0°C. The outer faces of
these outside plates are maintained at 0°C. Find the temperature of
the inner faces of the two plates and the temperature at the midpoint
of the inner slab 10 sec. after the plates have been put together. Given :
a — 0.2 c.g s. unit.
Hint: The boundary conditions are
u = 0
u = 0
u = f(x)
where /(x) is 0 when 0 < x < 1,
f(x) is 100 when 1 < x < 2,
f(x) is 0 when 2 < x < 3.
Hence,
when x — 0
when t = 0,
C3j>/\ •
L f(x) s
j f 2 ^ ™ • n<jrx j
sm ~r- dx = Ij 1 00 sin -rr- dx.
376 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §112
3. An insulated metal rod 1 m. long has its ends kept at 0°C., and its
initial temperature is 50°C. What is the temperature in the middle of
the rod at any subsequent time? Use k = 1.02, c = 0.06, and p = 9.6.
4. The faces of an infinite slab 10 cm. thick are kept at temp. 0°C.
If the initial temperature of the slab is 100°C., what is the state of
temperature at any subsequent time?
6. Let the rod of Prob. 3 have one of its ends kept at 0°C. and the
other at 10°C. If the initial temperature of the rod is 50°C., find the
temperature of the rod at any later time.
Hint: Let the ends of the rod be at x — 0 and x = 100; then the
conditions to be satisfied by the temperature function u(x, t) are as
follows: w(0, t) = 0, w(100, t) = 10, u(x, 0) = 50. Denote the solu-
tion of Prob. 3 by v(x, t) ; then if the function w(x, t) satisfies the
conditions *
Tt = a* d5' w(0' ° = °' ™(100' ° = 10' w(x> 0) = °'
u(x, t) — v(x, t) + w(x, t) will be the solution of the problem. Assume
the solution w(x, t) in the form w(x, t) = x/10 + <p(x, t), and deter-
mine the function <p(x, t).
6. Let a rod of length / have one of its ends x — 0 maintained at a
temperature u = 0, while the heat is dissipated from the other end
x — I according to the law
Let the initial temperature be u(x, 0) = f(x), where f(x) is a prescribed
function, Choose a particular solution of (112-1) in the form
e'-«-'0'< sin fix,
and show that the boundary conditions demand that
0 cos 01 = -hsm/31.
Write this transcendental equation in the form
tan 01= -Q,
and show that it has infinitely many positive real roots ]8i, 02, ' ' ' , 0n*
Hence, if
u(x,Q) = f(x)
then the solution has the form
n = l
§113
PARTIAL DIFFERENTIAL EQUATIONS
377
The functions sin/3na;, (n = 1, 2, • • • ), are easily shown to be orthog-
onal in the interval (0, I), so that the coefficients An in the solution are
given by the formula
i f(x) sin finx dx
A _ J°
/In — pi
I sin2 pnx dx
113. Vibration of a Membrane. Consider an elastic mem-
brane, of surface density p, which is under uniform tension T. By
definition the tension is said to be uniform if the force exerted
across a line of unit length in the plane of the membrane is
independent of the orientation of the line. It will be assumed
y
FIG. 100
that the plane of the membrane coincides with the £2/-plane of the
rectangular coordinate system and that the displacement of any
point of the membrane in the direction normal to the xy-plane is
denoted by z. Then a consideration of the forces acting upon
the element dA of the membrane (Fig. 100) leads to the equation
m*n
(113-1)
__
where c2 = T/p. The analysis leading to (113-1) is quite similar
to that used in deriving Eq. (108-2) for the vibrating string;
and, just as in Sec. 108, the underlying assumption here is that
the displacement z is not too great.
The solution of the problem of a vibrating membrane consists
of determining the function z = f(x, y, t), which satisfies the
differential equation (113-1) as well as the boundary and initial
conditions characteristic of the particular physical problem under
378 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §113
consideration. These remarks will be illustrated by a brief
treatment of the case in which the membrane is circular. In this
case the shape of the membrane suggests the use of the cylindrical
coordinate system in preference to the rectangular system. As
will be seen presently, the choice of cylindrical coordinates is
made because the boundary conditions assume particularly simple
forms in these coordinates.
The transformation of (113-1) can be accomplished readily
with the aid of the relations* connecting cylindrical coordinates
with rectangular, namely,
x = r cos By y = r sin 0, z = z
or
r = \fifi~+~y*, 0 = tan"1 -> z = z.
X
It will be necessary to express d2z/dx2 and d2z/dy2 in terms of the
derivatives of z with respect to r and 6. Nowf
dz = dz dr dz <90
dx ~ dr dx 68 dx
and
/ \ 2
d2r dz d2z i dO\ d26 dz d2z dr
. I VV I . V Vt* . ^ V ^ V'
a0
dx2 dr2 \dx/
But
dr x
ao:2 dr T a02 \dxj ] dx2 dd ] " dr dS dx
aa;
dx T/X* q: y
d2r y2
sin2 6 d26 2xy 2 sin 6 cos 6
dr2 (r2 4- ?/2V^
t/x {J, \ y )
Substituting these
d*z d*z cos* e +
r ' dx2 (x2 + y2)2 r2
values in the expression for d2z/dx2 gives
dz sin2 6 d2z sin2 0 dz 2 sin 0 cos 0
«
Similarly,
— „ 01 •»-» 2 fl 1.
ar r ' a02 r2 ' a0 r2
0 a2z ( — cos 0 sin
1 "ara0 r
dz cos2 0 a2z cos2 0 az ( — 2 sin 0 cos 0
')
~ o 10 olll V p
at/2 ar2
* See Sec. 56.
t See Sec. 39.
ar r a02 r2 . a0 r2
0 a2z /sin 0 cos
'Y
ar a0 \ r
/
§113 PARTIAL DIFFERENTIAL EQUATIONS 379
so that
— z + ^z = ~ 4- i — 4- 1 *?5
to2 + <ty2 ~ dr2 + r dr + r2 d02
and (113-1) can be written as
(H3-2) S =
dr
It was remarked in Sec. 104 that the solution of such an equa-
tion contains two arbitrary functions, and in order to make
the problem definite it is necessary to know the initial and
boundary conditions. Thus, suppose that the membrane is of
radius a and is fastened at the edges. Then it is evident that the
solution of (113-2),
z = F(r, 6, 0,
must satisfy the condition z = 0 when r = a, for all values of t.
If, moreover, the membrane is distorted initially into some
surface whose equation is a function of the radius only (that is, the
initial distortion is independent of 0), say z = /(r) when t = 0,
then it is clear that the subsequent motion will preserve the
circular symmetry and that the solution will be a function 6f
r and t only. These conditions alone are not sufficient for the
unique determination of the function
z = F(r, 0,
and it is necessary to specify the initial velocity of the membrane
in order to make the problem perfectly definite. If the mem-
brane is distorted and thereafter released from rest, then
~ = 0 when t = 0.
ot
Since the solution is assumed to be independent of 0, (113-2)
becomes
(113-3) ^ = <
and its solution satisfying the boundary and initial conditions
z — 0 when r = a,
(113-4) /=/(r) when* = 0,
-r- = 0 when t = 0
at
will be obtained by a method similar to that used in Sec. 111.
380 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §113
Assume that it is possible to express the solution of (113-3) as
the product of two functions, one of which is a function of r alone
whereas the other is a function of t alone. Then,
Substitution of this relation in (113-3) leads to
R ldR\
+ r dr)
dt*
or
(1135)
(il6~d)
Since the left-hand member of (113-5) is, by hypothesis, a func-
tion of t alone, whereas the right-hand member is a function of
r alone, each member must be equal to some constant, say — w2.
Hence, (113-5) can be written as
(113-6) + «*r = 0
and
,
dr2 r dr
where k = u/c.
Equation (113-6) is the familiar equation of simple harmonic
motion, and Eq (113-7) is easily reducible to the Bessel equation
by the substitution x = kr. Thus, if x — kr,
dR = dR dr = IdR
dx dr dx k dr'
d*R d (ldR\ I d*R dr 1 d*R
dx2 dx \kdrj k dr* dx k* dr*'
so that (113-7) assumes the form
d*R IdR
dx^+xd^+R==°>
which possesses the solution (see Sec. 100)
R = J0(x) = Jo(fcr).
Therefore,
z = RT = Jo(fer) sin co£
or
z = Jo(kr) cos ait.
§113 PARTIAL DIFFERENTIAL EQUATIONS 381
X
Since the last of the boundary conditions (113-4) requires
— == 0 when t = 0,
it is necessary to reject the solution involving sin co£. Further-
more, the first of these conditions demands that
2 = 0 when r = a,
so that
z = Jo(ka) cos co? = 0
for all values of t. This condition will be satisfied if the arbitrary
constant k is so chosen that Jo(fca) = 0. In other words, ka
UoOO
FIG. 101
must be a root of the Bessel function of order zero (Fig. 101); and
if the r*th root of Jo(kr) is denoted by
then
Since k = w/c, it follows that
co = knc.
Hence, a solution of (113-3) that satisfies two of the boundary
conditions (113-4) is given by
Jo(knr) cos knct.
The sum of any number of such solutions, each multiplied by an
arbitrary constant, will be a solution, so that
(113-8) z = 2^ AnJo(knr) cos knct
will be a formal solution of (113-3).
382 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §114
But when t = 0, the second of the boundary conditions
demands that z — f(r). Since (113-8) becomes, for t = 0,
00
(113-9) z = J A«J»(knr),
n — l
it follows that, if it is possible to choose the coefficients in the
series (113-9) so as to make
00
(113-10) AnJ<,(knr) = f(r),
then (113-7) will be the required formal solution of (113-3) which
satisfies all the conditions (113-4).
The problem of development of an arbitrary function in a
series of Bessel functions has been discussed in Sec. 101, where it
was indicated that a suitably restricted function /(r) can be
expanded in a series (113-10), where
114. Laplace's Equation. Let it be required to determine
the permanent temperatures within a solid sphere of radius unity
when one half of the surface of the sphere is kept at the constant
temperature 0°C. and the other half is kept at the constant
temperature 1°C.
From the discussion of Sec. 110, it is evident that the tempera-
ture u within the sphere must satisfy Laplace's equation
d2u d*u d*u __
dx* + dy* + fc*
The symmetry of the region within which the temperature is
sought suggests the use of spherical coordinates. If Laplace's
equation is transformed with the aid of the relations* (Fig. 102)
x = r sin 6 cos <p,
y = r sin 6 sin <p,
z = r cos 0}
in a manner similar to that employed in Sec. 113, the equation
becomes
* See Sec. 56.
§114
PARTIAL DIFFERENTIAL EQUATIONS
383
It is necessary to seek a solution of this equation that will satisfy
the initial conditions.
If the plane separating the unequally heated hemispheres is
chosen so that it coincides with
the xy-pl&ne of the coordinate
system and if the center of the
sphere is taken as the origin,
then it appears from symmetry
that it is necessary to find the
temperatures only for that por-
tion of the sphere which lies to
the right of the #2-plane (see
Fig. 102). Moreover, it is clear
that the temperatures will be
independent of <p, so that (114-1) FIG 102
becomes
The solution of (114-2) must be chosen so as to satisfy the bound-
ary conditions
(114-3)
u = 1 for 0 < 0 < ~ when r — 1,
u = 0 for -
< TT when r = 1.
In order to solve (114-2), assume that the solution
u = F(r, 6)
is expressible as the product of two functions, one of which is
independent of 6 and the other independent of r. Thus, let
u = /2(r)6(0).
The substitution of this expression in (114-2) leads to the two
ordinary differential equations
d\rR)
dr2
- a*R = 0
and
1 d
sin0^) + a*e = 0,
where a2 is an arbitrary constant.
384 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §114
The first of these equations can be expanded to read
2d2R , 0 dR 2P A
r2 -T-Y + 2r -^ -- a2.R = 0,
ar2 ar
which is an equation of the type treated in Sec. 97 and the linearly
independent solutions of which are
R = rm and R = l/rm+\
where
so that
a2 = m(m + 1).
If this value of a is substituted in (114-4), this equation becomes
The change of the independent variable 6 to x by means of
x = cos 6 leads to Legendre's equation
If m is an integer, particular solutions of this equation are the
Legendre polynomials
P(T\ _ p (P0u fi\
m\>") — -*• mv^-^-'O l/y,
and hence the particular solutions of (114-2) are
u = rmPm(cos 6),
_ Pm(cos 6)
lAi — ~r~: •
The second of these solutions evidently cannot be used, for it
becomes infinite when r — > 0. Therefore, it will be necessary to
build up the expression for the temperature u within the sphere
from terms of the type rmPm(cos 0), where m is a positive integer.
Consider the infinite series
00
(114-5) u = 2 Amr™Pm(cos 0),
each term of which satisfies (114-2). When r = 1, (114-5)
becomes
00
ra-0
§114 PARTIAL DIFFERENTIAL EQUATIONS 385
and, if it is possible to choose the undetermined constants Am in
such a way that (114-5) satisfies the boundary conditions (114-3),
then (114-5) will be the desired solution of the problem. ' Now,
it was indicated in Sec. 102 that a suitably restricted function
y = F(x)
can be expanded in the interval ( — 1, 1) in a series of Legendre
polynomials in the form
F(x) =
where the coefficients am are given by
(114-6) am = ^^ J^F(aOPm(z) dx.
In the problem under consideration,
u = f(o) = 1 for 0 < 6 < £>
2i
u = f(o) = 0 for I < 6 < TT,
so that the problem is equivalent to expanding F(x) as
F(x) = i AmPm(x),
where F(x) = 0 for -1 < x < 0, and F(x) = 1 for 0 < x < 1.
If formula (114-6) is used, it is readily found that the solution of
(114-2), which satisfies the initial conditions (114-3), is
+ 5-ri3^(coB») ---- -
PROBLEMS
1. Find the steady-state temperature in a circular plate of radius a
which has one half of its circumference at 0°C. and the other half at
nrc.
Hint: Use Laplace's equation for the plane in polar coordinates,
<Pu ,ldu l^d^u
6r* + r dr + r2d6* " U>
and assume that u = R(r)B(6) as in Sec. 114. Hence, show that the
386 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §115
physically possible solution is of the form
u = a0 + air cos 6 + a2r2 cos 26 + a3r3 cos 30 + • • •
+ bir sin 6 + 62r2 sin 26 + b^ sin 30 + • • • .
Determine the coefficients a, and 6t so as to satisfy the boundary
conditions.
2. Show that Laplace's equation in cylindrical coordinates is
d*u I du 1 dzu d*u _
dr* + r dr + r2 d<p* + dz2 ~
and in spherical coordinates is
d ( zdu\ . 1 d f . -du\ . 1 d*u
~*~ \ r i~ } + ~ — Q ~™ \ SHI 0 -^ J 4- . o /) ^™ 5 = 0.
dr\ dr / sin 6 00 \ SO / sin2 6 o<p*
3. Find the steady-state temperature at any point of a semicircular
plate of radius a, if the bounding diameter of the plate is kept at the
temperature 0°C. and the circumference is kept at the temperature
100°C.
Hint: Use Laplace's equation for the plane in polar coordinates, namely,
d*u 1 du 1 dzu _
dr2 + r dr + r2 d02 ~~
4. Outline the solution of the problem of the distribution of tem-
perature in a long cylinder whose surface is kept at the constant tem-
perature zero and whose initial temperature in the interior is unity.
115. Flow of Electricity in a Cable. A simple problem of
determining the distribution of current and voltage in an elec-
trical circuit, whose linear dimen-
----- oc+Ax---->\ . n xi_ x
. _____ x _.__>i sions are so small that one can
- 1 JL - T^ disregard the variation of the e m.f
I > B '<dwig the circuit, has been discussed
JE ^ in Sees. 90 and 91. This section
^ is concerned with the more compli-
cated problem of the flow of elec-
tricity in linear conductors (such as telephone wires or submarine
cables) in which the current may leak to earth.
Let a long imperfectly insulated cable (Fig. 103) carry an
electric current whose source is at A. The current is assumed
to flow to the receiving end at R through the load B and to
return through the ground. It is assumed that the leaks occur
along the entire length of the cable because of imperfections in
the insulating sheath. Let the distance, measured along the
§115 PARTIAL DIFFERENTIAL EQUATIONS 387
length of the cable, be denoted by x; then both voltage and
current will depend not only on the time t, but also on the dis-
tance x. Accordingly, the e.m.f. F (volts) and the current 7
(amperes) are functions of x and t. The resistance of the cable
will be denoted by R (ohms per mile) and the conductance from
sheath to ground by G (mhos per mile). It is known that the
cable acts as an electrostatic condenser, and the capacitance of the
cable to ground per unit length is assumed to be C (farads per
mile); the inductance per mile will be denoted by L (henrys
per mile).
Consider an element CD of the cable of length Ax. If the
e.m f. is F at C and F + AF at D, then the change in voltage
across the element Ax is produced by the resistance and the
inductance drops, so that one can write
AF = -
The negative sign signifies that the voltage is a decreasing func-
tion of x. Dividing through by A:r and passing to the limit as
Ax — > 0 gives the equation for the voltage,
(115-1) y- = -IR - L !»•
The decrease in current, on the other hand, is due to the
leakage and the action of the cable as a condenser. Hence, the
drop in current, A7, across the element Ax of the cable is
A7 = - VG Ax - ~ C Ax.
ot
so that
(115-2) — = — VG — C— •
Equations (115-1) and (115-2) are simultaneous partial
differential equations for the voltage and current. The voltage
F can be eliminated from these equations by differentiating
(115-2) with respect to x to obtain
d27 dV d2V
'dx2 ~~ "~ ~dx "" dx dt
Substituting for dV /dx from (115-1) gives
388 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §116
527
from which d2V/dx dt can be eliminated by using the expression
for d*V/dtdx obtained from the differentiation of (115-1).
Thus, one is led to
(115-3) -Lc
A similar calculation yields the equation for F, namely,
r)2F <92F r)F
(115-4) - LC - - (LG + RC)d-- KGV = 0,
which is identical in structure with (115-3).
In general, it is impossible to neglect the capacitance C of the
cable in practical applications of these equations to problems in
telephony and telegraphy, but the leakage G and the inductance
L, normally, are quite small. Neglecting the leakage arid induc-
tance effects yields the following equations:
(115-5) |£ = -IR,
dl dV
It is clear from (115-7) and (115-8) that the propagation of
voltage and current, in this case, is identical with the flow of
heat in rods.
In order to give an indication of the use of these equations,
consider a line / miles in length, and let the voltage at the source
A, under steady-state conditions, be 12 volt's and at the receiving
end R be 6 volts. At a certain instant t — 0, the receiving end is
grounded, so that its potential is reduced to zero, but the poten-
tial at the source is maintained at its constant value of 12 volts.
The problem is to determine the current and voltage in the line
subsequent to the grounding of the receiving end.
It follows that one must solve Eq. (115-8) subject to the follow-
ing boundary conditions :
§116 PARTIAL DIFFERENTIAL EQUATIONS 389
V = 12, at x = 0 for all t > 0,
7 = 0, at x = I for all t > 0.
In addition, it is necessary to specify the initial condition that
describes the distribution of voltage in the line at the time t = 0.
Now, prior to the grounding of the line, the voltage V is a func-
tion of x alone, so that (115-8) gives
the solution of which is
V = c\x + c%.
Since, prior to grounding, V = 12 at x = 0 and V = 6 at x = Z,
it follows that c\ = — 6/7 and c2 = 12, so that
fir
V = - y + 12 at t = 0.
Accordingly, it is necessary to find the solution of Eq. (115-8)
subject to the following initial and boundary conditions:
F(0, 0 = 12, V(l, t) = 0,
V(x, 0) = - «* + 12.
^
A reference to Sec. 112 shows that the mathematical formula-
tion of this problem is similar to that of the problem of heat flow
in a rod, except for the difference in the formulation of the end
conditions. * Now, the voltage V(x, t) in the line, subsequent to
the grounding, can be thought of as being made up of a steady-
state distribution Vs(x) and the transient voltage VT(X, t), which
decreases rapidly with the time. Thus,
V(X, t) EEE VS(X) + VT(X, t).
After the line has been grounded, the voltage at the ends of
the line must satisfy the following conditions:
F(0, 0 = 12 and V(l, t} = 0.
It was noted above that the steady-state distribution of voltage
is a linear function of x\ and since after the lapse of some time t
the transient effects will not be felt, it follows that
Va(x) = - y x + 12.
* See, however, Prob. 5, Sec. 112.
390 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §115
Thus,
(115-10) V(x, t) = - y x + 12 + VT(x, t).
The boundary conditions to be satisfied by the transient
voltage VT(X, t) can now be determined from (115-9). Thus,
7(0, t) = 12 = 12 + Fr(0, 0,
V(l, 0=0= Fr(Z, 0,
fir 197*
F(«, 0) = - y + 12 = - i~ + 12 + Fr(*, 0).
Hence, the function VT(x, t) satisfies the following initial and
boundary conditions:
Fr(0, 0 = VT(l, t) = 0,
Since it is obvious from (115-10) that VT(X, t) satisfies (115-8),
it becomes clear that the determination of the transient voltage
VT(X, t) is identical with the problem of determining the dis-
tribution of the temperature in a rod when the initial distribution
is the linear function fix /I. Referring to the solution (112-5)
and setting a2 = \/(RC) give
(,\ ^^_ i / & I \j • flTTX 7 \ T77H ~~i~ I ' • TlTTX
^ 0 = >, I 7 I 7 x sm ~~T dx )e ^ J sm -T—
A x?/2 r° • n*xj\
X, t) = ^ ( j I j x sin -y- dx 1
Therefore, the problem of determining the distribution of voltage
is solved.
The magnitude of the current in the line is obtained from
(115-5). It is left as an exercise for the reader to calculate
the expression for the current /. It is easy to see that the term-
by-term differentiation of the series for VT(X, t) is valid for all
values of t > 0.
From the discussion of this problem, it is clear that the deter-
mination of the temperature of a rod whose ends are kept at
different fixed temperatures and whose initial temperature is a
function of the distance along the rod can be effected in a similar
way.
PROBLEMS
1. On the assumption that the length I of the line in Sec. 115 is 120
miles, R — 0.1 ohm per mile, and (7 = 2- 10~8 farad per mile, find the
COMPLEX VARIABLE 441
called the plane of a complex variable, the x-axis is called the
real axis, and the 2/-axis is called the imaginary axis.
If v vanishes, then
z — u + 0 - i = u
is a number corresponding to some point on the real axis. Accord-
ingly, this mode of representation of complex numbers (due to
Gauss and Argand) includes as a special case the usual way of
representing real numbers on the number axis.
The equality of two complex numbers,
a + ib = c + id,
is interpreted to be equivalent to the two equations
a — c and b = d.
In particular, a + ib = 0 is true if, and only if , a = 0 and 6 = 0.
If the polar coordinates of the point (u, v) (Fig. 123) are
(r, 6), then
u = r cos 0 and v = r sin 0
so that
r = \/u2 + *>2 and 0 = tan"1 -
11
The number r is called the modulus, or absolute value, and 6
is called the argument, or amplitude, of the complex number
z = u + iv. It is clear that the argument of a complex number
is not unique ; and if one writes it as 6 + 2&?r, where — TT < 6 ^ TT
and fc = 0, ±1, ±2, • • • , then 6 is called the principal argu-
ment of z. The modulus of the complex number z is frequently
denoted by using absolute value signs, so that
r = \z\ = \u + iv\ = v^2 + ^2>
and the argument 6 is denoted by the symbol
6 = arg z.
The student is assumed to be familiar with the fundamental
algebraic operations on complex numbers, and these will not be,
entered upon in detail here. It should be recalled that
21 + 22 = (xi + iyi) + (x2 + iy*) = (xi + z2) + i(y\ + 2/2),
21 • 22 = G&i + iy\) - (x2
21 i i _ i2 i2 , z-
22 z2 + **2/2 ^1 +2/1 ^i + 2/1
provided that j^l =
442 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §133
It follows from the polar mode of representation that
z\ * 82 = ri(cos 0i + i sin 0i) • r2(cos 02 + i sin 02)
= rir2[cos (0i + 02) + i sin (0i + 02)];
that is, the modulus of the product is equal to the product of the
moduli and the argument of the product is equal to the sum of the
arguments. Moreover,
zi ri (cos 0i + i sin 0i)
- — — ~ — ~
i r ,
= ~~ Lcos (0i ""
^ sm
/>
— 02
— —7 - 2 — i — ~ — n~\ ~~
z2 r2 (cos 02 + 1 sin 02) r2
so that ^e modulus of the quotient is the quotient of the moduli and
the argument of the quotient is obtained by subtracting the argument
FIG. 124.
of the denominator from that of the numerator. If n is a positive
integer, one obtains the formulas of De Moivre, namely,
\/z = {r[cos (0 + 2kw) + i sin (0 +
so that
0 + 2/C7T
>s
R(cos <p + i sin <
0 -
0 + 2kir\
n )
(* = 0, 1, 2,
n-1).
This last formula can be illustrated by finding the expressions
for the cube roots of z = 1 — i. Since u = 1 and 2; = —1, it
follows that r = -\/2 and 0 = tan-1 ( -y- )• Hence
§133
COMPLEX VARIABLE
443
-f + 2*r
sin
Assigning k the values 0, 1, and 2 gives the values of the three
roots as (Fig. 124)
«- -
and
*1 = ^[
6/0 I 7?T
22 = V2^C08 ^
23 = A/2 ( COS —
. . .
+ i sin
in (" s))
in T£ I
i^/
inj)
The following important inequalities will be recalled, for they
are used frequently in this
chapter.
(133-1) \zi + z*\
£ N + N,
that is, the modulus of the sum
is less than or equal to the sum
of the moduli. This follows
at once upon observing (Fig.
125) that the sum of two FlG- 125'
sides of the triangle is not less than the third side.
(133-2) \zi + z2| ^ M - M,
that is, the modulus of the sum is greater than or equal to the differ-
ence of the moduli. This follows
from the fact that the length of
one side of a triangle is not less
than the difference of the other two
sides.
(133-3) |2i| — \z%\
FIG. 126.
This follows from Fig. 126. •
PROBLEMS
1. Find the modulus and argument of
(a) 1 + i V3, (6) 2 + 2i,
(c) (1 +
- i).
444 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §134
7T IT.
2. If z\ = 3e* and zz — e ® , find 3i • z2 and 21/22. Illustrate the
results graphically.
3. Under what conditions does one have the following relations?
(a) \z\ -j- 22! = \z\\ -f- \Zz\t
(&) |#1 ~f" #2! == |^l| — l^l-
4. Setting 2 = r(cos 6 -\- i sin ^), show, with the aid of the formula
of De Moivre zn = rn(cos nd + i sin n#), that
cos 2^ = cos2 6 — sin2 0 and sin 20 = 2 sin 0 cos 0.
6. Find all the fifth roots of unity, and represent them graphically.
6. Find all the values of \Xl + i and \/i, and represent them
graphically.
7. Find all the roots of the equation zn — 1 = 0.
8. Write the following complex numbers in the form a + bi:
(a)
(b)
(c) (1 - V301;
9. Express the following functions in the form u +
1 (c) 2;2 - z + 1 ;
(a)
(6)
10. The conjugate of a complex number a + ib is defined as a — t6.
Prove that
(a) The conjugate of the product of two complex numbers is equal
to the product of the conjugates of the complex numbers.
(6) The conjugate of the quotient of two complex numbers is equal
to the quotient of the conjugates of the complex numbers.
134. Elementary Functions of a Complex Variable. A com-
plex quantity z = x + iy, where x and y are real variables, is
called a complex variable. If the assignment of values to z
determines corresponding values of some expression /(z), then
f(z) is said to be & function of the complex variable z. For example,
if
§134 COMPLEX VARIABLE 445
the values of f(z) can be determined by recalling that, if z = x
+ iyy then
So long as the functions under consideration involve only the
operations of addition, subtraction, multiplication, division, and
root extraction, the discussion of Sec. 133 provides methods of
determining the values of these functions when arbitrary values
are assigned to z Thus, if /(z) is any rational function of z,
that is, the quotient of two polynomials so that
ft \ - apz" + a\zn~l + • * • + an
J(Z) ~ ~"'-1 • • 7
there is no difficulty in ascertaining its values. The discussion
permits one to ascribe a meaning even to such an expression as
1 *
— i V^-zr
For, if z — x + iy, then
2ixy
= [r(cos 9 + i sin 0
where
r = V(x* - y2 - I)2 + 4zV and 0 = tan 2 — -
x y 1
Applying De Moivre's formula gives
V?'"! = r^[cos H(« + 2/C7r) + i sin %(0 + 2fcx)], (fc = 0, 1),
and therefore,
19 + 2/C7T
. .
* sin -_ , (t „ 0| 1).
Matters become somewhat more involved when it is necessary
to define transcendental functions of z such as
ez, sin z, log 2, etc.
It is evident that it is desirable to define these functions so that
they will include as special cases the corresponding functions
of the real variable x. It was indicated in Sec. 73 that the series
446 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §134
which converges for all real values of x, can be used to define
the function ezy where z = x + iy, so that
Kv2 v* \
i — ^ I- — — ... 1
2! ^ 4! /
/ y3 V5 M
+ H # ~~ 3J + 51 "" ' ' ' ) = ez(cos y + i sin i/).
Also, from Sec. 73,
ev» _L
COS I/ =
sin y —
2
evl —
These formulas lead one to define the trigonometric functions
for a complex variable z as
ezl + e~zl . ezl — e~zl
cos z — ~ -J sm z =
2 ' ^A~ 2i
sin z , cos z
tan z = > cot z = — — >
cos z sin 2
1 1
sec z = > esc z =
cos z sin z
It can be easily verified by the reader that these definitions
permit one to use the usual relations between these functions,
so that, for example,
g«! • QZI = ezi+*2,
sin2 z + cos2 2=1,
sin (zi + #2) = sin z\ cos #2 + cos z\ sin 22.
The logarithm of a complex number z is defined in the same
way as in the real variable analysis. Thus, if
w = log 2,
then ~~~~ "*
z = e".
Setting w = u + iv gives
2 — 6u+it> = eu(cos v + i sin v).
On the other hand, z can be written as
z = x + iy = r(cos 6 + i sin 0).
§134 COMPLEX VARIABLE 447
Therefore,
r(cos 6 + i sin 0) == ew(cos v + i sin v),
which gives
e» = r, t; = 0 + 2&7r, (fc = 0, ± 1, ±2, ± • • • ).
Then, since u and r are real, u = log r, so that
(134-1) w = u + iv = log z = log r + (0 + 2kir)i.
Hence, the logarithm of a complex number has infinitely many
values, corresponding to the different choices of the argument of
the complex number. Setting k = 0, one obtains the principal
argument of log z, if it is assumed that — TT < 6 < TT.
It is obvious that (134-1) provides a suitable definition of
log z for all values of z with the exception of z = 0, for which
log z is undefined.
The definition (134-1) permits one to interpret the complex
power w of a complex variable z by means of the formula
gW __ gW log Z •
and since log z is an infinitely many-valued function, it follows
that, in general, zw likewise has infinitely many values.
PROBLEMS
1. Verify the formulas
(a) e*i • ez* = ezi+**;
(b) sin2 z + cos2 z = 1 ;
-. (c) cos (zi + #2) = cos zi cos Zz — sin z\ sin 32;
/0(d) cos iz — cosh z\
A (e) sin iz = i sinh z:
"j } '*•
2. Represent graphically the complex numbers defined by the
following:
(a) logi; (d) i1',
(b) log(-l); _ (e) e*<.
(c) log (I - V3i);
3. Show that
, . t I e*< - I
(a) tan«-T?ir?1;
™ * -«^±1
(6) cot 2 = t
448 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §135
4. Express tan z in the form u + w.
6. Express sin z in the form u + iv.
6. If a and 6 are real integers, show that
(re0t)a+&t - rae-be[cos (ad + b log r) + i sin (ad + b log r)].
7. Write in the form r(cos 6 + i sin 0)
(a) (1 + 0*; (d) Is
(b) (1 - O1-; W 21+».
(c) t'1-');
136. Properties of Functions of a Complex Variable. Let
w = /(z) denote some functional relationship connecting w with
z. If z is replaced by x + iy, w can be written as
w = f(x + iy) = w(x, y) + iv(x, y),
where u(x, y) and v(x, y) are real functions of the variables
x and y. As an example, one may consider the simple function
w = z2 = (x + iy)2 = x2 — yz + 2ixy.
If x and ?/ are allowed to approach the values XQ and y0, respec-
tively, then it is said that the complex variable z = x + iy
approaches z0 = XQ + iyo. Thus the statement
z — » z0, or a: + iy — > o:0 + iyo,
is equivalent to the two statements
x — » x0 and y — » y0-
Since /(^0) is, in general, a complex number, one extends the
definition of continuity in the following way: The function f(z)
is said to be continuous at the point z = z0 provided that
(135-1) /(z) — »/(ZQ) when z — > z0.
Since /(z) = u(x, y) + iv(x, y) and /(z0) = U(XQ, yQ) + iv(x0, 2/0),
the statement (135-1) implies the continuity of the functions
u(x, y) and v(x, y). If the function /(z) is continuous at every
point of some region R in the z-plane, then /(z) is said to be
continuous in the region R.
The complex quantities z and w can be represented on separate
complex planes, which will be called the z-plane and the w-plane,
respectively. Thus the functional relationship w = /(z) sets
up a correspondence between the points (x, y) of the z-plane and
the points (u, v) of the w-plane (Figs. 127 and 128).
§135
COMPLEX VARIABLE
449
If the variable z = x + iy acquires an increment Az, then
(Fig. 127)
z + Az = (x + Aor) + i(y + Ay)
and
Az — Ax + iky.
The change in w = /(z), which corresponds to the change Az in
z, can be denoted by Aw (Fig. 128), and one defines the derivative
of w with respect to z to be the f unction /'(z) such that
/(z + Az) - /(z)
(135-2)
/'(z) = lim
Az->0
Az
where the limit must exist and be independent of the mode of
approach of Az to zero.
v i
z- plane w- plane
w+Aw
FIG. 127.
FIG. 128.
It should be noted that this requirement, that the limit of
the difference quotient have the same value no matter how Az
is allowed to approach zero, narrows down greatly the class of
functions of a complex variable that possess derivatives. Thus,
consider the point P in the z-plane that corresponds to z = x + iy,
and let Q be determined by z + Az = (x + Az) + i(y + A£/).
In allowing the point Q to approach P, one can choose any one
of infinitely many paths joining Q with P, and the definition
(135-2) demands that the limit /'(z) be the same regardless of
which one of the paths is chosen.
, -Let it be assumed for the moment that w = /(z) has a unique*
derivative at the point P; then
* It is assumed throughout that we are concerned with single-valued
functions; hence, the discussion of the derivatives of such functions as
Vl — 2, for example, is restricted to a study of one of the branches of the
function.
450 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §135
(135-3) /'(z) = lim ^ " 2z — '
where
f(z) = u(x, y) + iv(x, y).
If Q is to approach P along a straight line parallel to the z-axis,
then Ay = 0, kz = Az, and
dz ~dx dx dx
On the other hand, if Q approaches P along a line parallel to the
y-axis, then Ax = 0, Az = i Ay, and
dz
Since the derivative is assumed to exist, (135-4) and (135-5)
require that the functions u(x, y) and v(x, y) satisfy the conditions
n . ^ du dv dv du
(IdO-O) — = —)
7 dx ^y
These are known as the Cauchy-Riemann differential equations,
and the foregoing discussion proves the necessity ol the condi-
tion (135-6) if f(z) = u(x, y) + iv(x, y) is to possess a unique
derivative.
In order to show that the conditions (135-6) are sufficient for
the existence of the unique derivative/' (2), one must suppose that
the functions u(x, y) and v(x, y) possess continuous partial
derivatives. *
It is not difficult to show that the usual formulas for the differ-
entiation of the elementary functions of a real variable remain
valid, so that, for example,
dzn n . de* d sin z
— = nzn~~l, -j- = c*, — -5 — = cos z, etc.
dz ' dz ' dz '
As an illustration of the application of the formulas (135-4) anvi
(135-5), consider the calculation of the derivative of
y) = 0* as ex+tv
or
w = w + iv = 6x(cos y + f sin y).
* Only the existence of these derivatives was required in the proof of the
necessity. See references at the end of Sec. 141.
COMPLEX VARIABLE 451
Here, u « e* cos t/, v = e* sin y, and it follows that
du du
•fa = e* cos i/, g£ = -e* sin y,
av . aw
gj - e- sin 2/, g£ = «• cos y.
Since Eqs. (135-6) are satisfied and the partial derivatives are con-
tinuous, dw/dz can be calculated with the aid of either (135-4) or (135-5).
Then,
dw ...
-T- = e* cos y + ie* sin y = e*.
The functions of a complex variable z that possess derivatives
are called analytic or holomorphic. * A point at which an analytic
function ceases to have a derivative is called a singular point.
It is possible to provef that, if f(z) is analytic in some region R
of the z-plane, then not only the first partial derivatives of u and
v exist throughout the region R, but also those of all higher orders.
This last statement leads to an important consequence of Eqs.
(135-6). Differentiating (135-6) gives
i
<W ^
and adding gives
Similarly, one obtains
Hence, the real and imaginary parts of an analytic function satisfy
Laplace's equation.
On the other hand, if a function u(x, y) satisfying Laplace's
equation is given, one can construct an analytic function f(z)
whose real part is u. Multiplying the first of Eqs. (135-6) by
dy and the second by dx and adding give
, dv , . dv , du , . du j
dv = — dx + T- dy = — — dx + — dy.
dx dy dy dx
Then, since du/dy and du/dx are known,
(135-7) v(x, y) = f ^ ( - ^ dx + g
J(xo,yo) \ ^2/ OX
* The term regular is also used.
f See Sec. 140.
452 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §135
where the line integral (135-7) will not be ambiguous if it is
independent of the path joining some fixed point (XQ, t/0) to
the point (x, y). Applying the conditions for the independence
of the path, * namely,
A ( - **\ = A (<*y
fy \ ty) " fa W
gives
to"2 "*" dy* ~ U'
which is precisely the condition assumed to be satisfied by u(x, y}.
Since the line integral (135-7) depends on the choice of the point
(x0, 2/0), it is clear that the function v(x, y) is determined to
within an arbitrary real C9nstant, and hence the function
f(z) = u + iv is determined save for a pure imaginary additive
constant.
It may be further remarked that the function v(x, y) may
turn out to be multiple-valued (if the region of integration is not
simply connected) even though u(x, y) is single-valued. The
connection of analytic functions with Laplace's equation is one
of the principal reasons for the great importance of the theory of
functions of a complex variable in applied mathematics, f
PROBLEMS
1. Determine which of the following functions are analytic functions
of the variable z = x + iy:
(a) x — iy;
(b) x* ~y* + 2ixy,
(c) Y2 log (z2 + y2) + i tan-1 (y/x);
2. Verify the following formulas:
, . d(cos z)
(a) — -— = - sin 2;
, x z) _
(c) dz = sec2 z:
K See Sec. 63.
t See, in this connection, Sees. 66, 111, and 130.
§136
COMPLEX VARIABLE
1
453
3. Find a function w such that w = u + iv is analytic if
(a) u = x* - y*]
(c) t* = x\
(d) u = log
(e) -u = cosh
4. Prove that
cos x.
);
(a) sinh z = K(e* ~ e~*) is analytic;
(6) cos (z + 2/C7T) = cos z, (/c = 0, ±1, ±2,
(c) sinh (2 + 2ikir) = sinh 2, (A; = 0, ±1, ±2, • • • );
(d) log 21 z2 = log 0i + log z2 ;
(e) log za = a log 2, where a is a complex number.
5. Show how to construct an analytic function /(z) = u(x, y)
+ iv (x, y) if v(a:, 2/) is given, and construct /(z) if v = 3a?22/ — ?/3.
6. An incompressible fluid flowing over the x7/-plane has the velocity
potential <£ = x1 — y-. Find a stream function ^.
7. Referring to Prob. 6, what is the velocity potential if the stream
function is
— ?3?
3x'2y — y
136. Integration of Complex Functions.
defined by the parametric equations y
Let C
x
x =
where <p and ^ are real differentiate
functions of the real variable t. Con-
sider a continuous (but not necessarily _
analytic) f unction /(«), of the complex
variable z = x + iy, defined at all points FIG. 129.
of C. Divide the curve C into n parts by inserting the points
Po, PI, ' ' • , Pn-i, Pn, where P0 coincides with the initial
point 2o of the curve and Pn with the end point zn (Fig. 129).
Let ft be any point on the arc of the curve joining JK*-i with Pt,
and form the sum
454 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §136
The limit* of this sum as n — > «> in such a way that each element
of arc Pt_iPt approaches zero is called the line integral of f(z)
along the contour C, that 'is,
(136-1) (j(z) dz = lim
*'c n— » »
The fact that this integral exists follows at once from the
existence of the real line integrals into which (136-1) can be
transformed. Indeed, separating f(z) into real and imaginary
parts as
/(«) = u(x, y) + iv(x, y)
and noting that dz = dx + i dy give
(136-2) fcf(z) dz = fc(udx-v dy) + i fc (v dx + u dy).
Thus, the evaluation of the line integral of a complex function
can be reduced to the evaluation of two line integrals of real
functions. It follows directly from the properties of real line
integrals that the integral of the sum of two continuous complex
functions is equal to the sum of the integrals, that a constant
can be taken outside the integral sign, and that the reversal of
the 'direction of integration merely changes the sign of the
integral.
It follows from (136-1), upon noting that the modulus of the
sum is not greater than the sum of the moduli, that
\fcf(z)dz < fc |/(s) | • \dz\.
If, along C, the modulus of f(z) does not exceed in value some
positive number M, then
(136-3) c /(*) dz ^ M c \dz\ = M c \dx + i dy\ = M c ds = ML,
where L is the length of C.
* The precise meaning of the symbol lim in (136-1) is the following: Con-
sider any particular mode of subdivision of the arc into n\ parts and denote
the maximum value of |z, — 2,_i( in this subdivision by Si, and let SHl stand
ni
for S /(f») fe ~ 2t-i). A new sum, corresponding to the subdivision of the
»-l
arc into n2 parts, is denoted by *S>n2; and the maximum value of z» — z,_i|
in this new subdivision is 52, etc. In this way, one forms a sequence of
numbers £Ul, £n2, • • • , Snm, • • • in which the numbers nm are assumed to
increase indefinitely in such a way that the 5» — * 0.
§138 COMPLEX VARIABLE 455
137. Cauchy's Integral Theorem. The discussion of the
preceding section involved no assumption of the analyticity of
the function /(z) and is applicable to any continuous complex
function, such as for example f(z) = z = x — iy, in which event
dz
If the integral (136-1) is to be independent of the path, then
it immediately follows from (136-2) that
du _____ dv_ eto _ du
~dy ~~ "" Hix fy ~ dx
Thus, the conditions that the integral of a complex function f(z)
be independent of the path are precisely the Cauchy-Riemann con-
ditions; in other words, the function f(z) must be analytic.
Now, let R be any region of the 2-plane in which f(z) is analytic,
and let C be a simple closed curve lying entirely within B; then
it follows from the properties of line integrals that* the following
important theorem holds:
CAUCHY'S INTEGRAL THEOREM. // f(z) is analytic within
and on a simple closed contour C, then fcf(z) dz = 0.
It should be noted carefully that the theorem has been estab-
lished essentially with the aid of Green's theorem, which requires
not only the continuity of the functions u and v but also the
continuity of the derivatives. Thus, the proof given above
implies not merely the existence of f'(z) but its continuity as
well.f It is possible to establish the validity of Cauchy's
theorem under the sole hypothesis that/' (z) exists and then prove
that the existence of the first derivative implies the existence
of derivatives of all orders. Accordingly, the proof given above
imposes no practical limitation on the applicability of the
theorem.
138. Extension of Cauchy's Theorem. In establishing
Cauchy's theorem in Sec. 137, it was assumed that the curve C
is a simple closed curve, so that the region bounded by C is
simply connected. It is easy to extend the theorem of Cauchy
to multiply connected regions in a manner indicated in Sec. 64.
Thus, consider a doubly connected region (Fig. 130) bounded
* See Sec. 63.
t See (135-4) and (135-5).
456 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §138
by the closed contours C\ and C2, where C2 lies entirely within Ci.
It will be assumed that the function f(z) is analytic in the region
exterior to C2 and interior to C\ and analytic on C2 and Ci. The
requirement of analyticity on Ci and C2 implies that the function
f(z) is analytic in an extended region (indicated by the dotted
curves KI and Jf£2) that contains the curves Ci and C2.
If some point A of the curve Ci is joined with a point B of C2
by a crosscut A B, then the region becomes simply connected and
the theorem of Cauchy is applicable. Integrating in the positive
direction gives
(138-1) C f(z)dz+ f f(z)dz + f f(z)dz+ f f(z) dz = 0,
JAPAO JAB JBQB&J JBA
where the subscripts on the integrals indicate the directions of
integration along Ci, the crosscut AB, and C2. Since the second
x. .^ and the fourth integrals in (138-1)
are calculated over the same path
in opposite directions, their sum is
zero and one has
where the integral along C\ is trav-
ersed in the counterclockwise direc-
FlQ' 13° tion and that along C2 in the
clockwise direction. Changing the order of integration in the
second integral in (138-2) gives*
This important result can be extended in an obvious way to
multiply connected regions bounded by several contours, to yield
the following valuable theorem.
THEOREM. // the function f(z) is analytic in a multiply con-
nected region bounded by the exterior contour C and the interior
contours Ci, C2, • • • , Cn, then the integral over the exterior contour
C is equal to the sum of the integrals over the interior contours
Ci, Ctj • • • , Cn- It is assumed, of course, that the integration over
* See Sec. 64.
§139
COMPLEX VARIABLE
457
all the contours is performed in the same direction and that f(z) is
analytic on all the contours.
139. The Fundamental Theorem of Integral Calculus. Let /(«)
be analytic in some simply connected region R, and let the curve
C join two points PO and P of R
(Fig. 131). The coordinates of P0
and P will be determined by the
complex numbers 20 and z. Now
consider the function F(z) defined
by the formula F(z) = /*,/(«) dz.
The function F(z) will not depend
upon the path j oining ZQ with z so long
as these points lie entirely within R.
Forming the difference quotient
gives
FIG. 131.
F(z
-F(z) _
/CO <fo -
r ^
f(z)dz\
Jzo J
In order to avoid the confusion that may occur if the variable z
appears in the limits and also as the variable of integration,
denote the latter by f , so that
F(z + &z) - F(z) = J
Az ~ A
(139-1)
/«•) rfr
snce
Now if
(139-2)
'' df = Az.
1 /»z
lim T- I
Az-»0 A2 Jz
458 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §139
then it follows from (139-1) that
dF(z)
dz
'/GO-
In order to prove that (139-2) holds, one merely has to make use
of (136-3) and note that max |/(f) - f(z)\ -> 0 as Az -» 0.
Any function F\(z) such that
dz J^ J
is called a primitive or an indefinite integral of /(«), and it is easy
to show that if F\(z) and ^2(2) are any two indefinite integrals of
f(z) then they can differ only by a constant.* Hence, if F\(z)
is an indefinite integral of f(z), it follows that
= f*f(z) dz = Fl(z) + C.
*fZO
In order to evaluate the constant C, set z = 20; then, since
(z) dz = Q,C = Ffa). Thus
(139-3) F(z) = f*f(z) dz =
JZQ
The statement embodied in (139-3) establishes the connection
between line and indefinite integrals and is called the fundamental
theorem of integral calculus because of its importance in the
evaluation of line integrals. It states that the value of the line
integral of an analytic function is equal to the difference in the
values of the primitive at the end points of the path of integration.
As an example consider
\ iri
6z fJy — pt — oie\ __ 1 — — 9
(*<& — t/ — o • JL — ~ £i»
This integral can also be evaluated by recalling that
* Proof: Since Fi'(z) = Fj(z) = /(«), it is evident that
Fi'(z) - F2'(z) = d(Fi - F2)/dz 9 dG/dz = 0.
But if dG/dz = 0, this means that G'(z) = ^ + i ~ = ^ - t — - 0, so
dx dx dy dy
,, . du dv dU dv rt j jj . j j j
that r— = -r-=s:"^~=:T~:=I0> an(l u an(l v do not depend on x and y.
dx dx dy dy
§139
Then,
COMPLEX VARIABLE
459
r/*(0,ir)
e* dz = I (e* cos y + ie* sin y)(dx + i dy)
/*(0,7r)
= I (e* cos ydx - e* sin y dy)
/•(o,»)
+ t J(0 (ex sin ydx + e* cos y dy).
As a more interesting example, consider
where n is an integer and the integral is evaluated over some curve
joining z0 and z. If n 9^ — 1, an indefinite integral is
+ 1
— a) n is analytic throughout the
For n > 0, the integrand /(z) =
finite z-plane and hence
(139-4) F (z - a)" dz = — r-r [(z - a)"+1 - (z0 - a)"+ 1].
Jzo n -f- i
If the variable point z is allowed to start from z0 and move along some
closed contour C back to z0, then
(*-
= 0.
Of course, the latter result could have been obtained directly from
Cauchy's integral theorem.
Suppose next that n < — 1 and that the path of integration does not
pass through the point a. If the point a is outside the closed contour
(7, then the integrand is analytic and it
follows at once from Cauchy's integral
theorem that
(z - a)" dz = 0.
Suppose now that the point a is within the
contour C. Delete the point a by enclos-
ing it in a small circle of radius p, and con-
sider the simply connected region R shown
in Fig. 132. Then, so long as n ^ — 1?
the single-valued function /(z) = (z — a)n is analytic in R and (139-4)
is applicable to any curve C joining z0 and z in R. Now if z is allowed
to approach ZQ, then it follows from the right-hand member of (139-4)
that
(z - a)* dz = 0 for n ^ - 1.
FIG. 132.
\c
460 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §139
There remains to be investigated the case when n — — 1. For any
path C not containing z = a, one obtains
(139-5)
dz .
(z-a)
z — a
Z Irvrr
a
n\
_ — log
— log _
a
a
log
ZQ — d
z — a
+ i arg _
a
log
ZQ — a
+ i arg (z
a) i arg (#o ~
a).
Now if the point z starts from z0 and describes a closed path C in such
a way that a is within the contour, then the argument of z — a changes
by 2?r, and therefore
./• /It
= 2iri.
— a
If a is outside the contour, then (z — a)~l is analytic within and on C
and hence the line integral is zero by
Cauchy's theorem.
A different mode of evaluating the
integral
where n is an integer greater than unity
>A: and C is a closed contour, will be given
next. If the point a is outside CV then
IG* 133' the value of the integral is zero by
Cauchy's theorem. Accordingly, consider the case when a is inside C.
Draw a circle 7 of radius p about the point a (Fig. 133) and, since the
integrand is analytic in the region exterior to 7 and interior to C, it
follows from the theorem of Sec. 138 that
Jc (z - a)~n dz = J (z - a)~n dz.
But z — a — pe°l and dz = ipedi dd on 7, so that
dz /*27r ipet& dO
/* dz fir ipel dO i r2jr
I i -r = I ~ 5- = r I
Jc (z — a)n Jo pne*n9 pn~l Jo
i
^~i
= o,
dd
\lU7*\.
This is the same result as that obtained above by a different method.
The reader should apply the latter method to show that, if a is inside C,
then J (z — a)-1 dz
2iri.
§140 COMPLEX VARIABLE 461
PROBLEMS
1. Show that J*o zdz = %(z2 — 202) for all paths joining 20 with z.
2. Evaluate the integral J c (z — a)"1 dz, where C is a simple closed
curve and a is interior to C, by expressing it as a sum of two real line
integrals over C.
Hint: Set z — a — pe9*; then dz = e*l(dp + i pdd).
3. Evaluate J c z~2 dz where the path C is the upper half of the unit
circle whose center is at the origin. What is the value of this integral
if the path is the lower half of the circle?
4. Evaluate J c z-1 dz, where C is the path of Prob. 3.
6. Evaluate fc (z2 - 2z + 1) dz, where C is the circle x1 + y* = 2.
6. Discuss the integral J G (z + l)/z2 dz, where C is a path enclosing
the origin.
7. What is the value of the integral J c (1 + 22)"1 dz, where C is the
circle x2 + y2 = 9?
8. Discuss Prob. 7 by noting that . , ^ = 2"' ( " -. '' 4^i anc^
evaluating the integrals over the unit circles whose centers are at
z — i and z — —i. Note the theorem of Sec. 138.
140. Cauchy's Integral Formula. The remarkable formula
that is derived in this section permits one to calculate the value
of an analytic function f(z), at any interior point of the region
bounded by a simple closed curve C, from the prescribed bound-
ary values of f(z) on C.
Let f(z) be analytic throughout the region R enclosed by a
simple closed curve C and also on the curve C. If a is some
point interior to the region R (see Fig. 133), then the function
(140-1)
z — a
is analytic throughout the region R, with the possible exception
of the point z = a, where the denominator of (140-1) vanishes.
If the point a is excluded from the region R by a circle 7 of radius
p and with center at a, then (140-1) is analytic throughout the
region exterior to 7 and interior to C, and it follows from Cauchy's
integral theorem that
f M.
Jco * - <*>
f -M-&= f -1^-dz.
Jco * - a JyO z - a
462 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §140
or
(l40-2)
The integral in the right-hand member of (140-2) can be
written as
(140-3) f I®- d* - f M^ dz + /(a) f J*L.
JT* - a JY 2 - a J7* - a
It was demonstrated in Sec. 139 that
f dz
I -- =
J7 z - a
. and it will be shown next that the first integral in the right-hand
member of (140-3) has the value zero. Set z — a = peBl;
then, so long as z is on 7, dz = ipeld dO,
and hence
f f(z) ~ /(^ <fe = i f
Jy Z a Jy
(140-4) ~ <fe = i [/(*) - /(a)] cW.
Jy Z a Jy
If the maximum of \f(z) — /(a) | is denoted by M, then it follows
from Sec. 136 that
(140-5)
z — a
2*
o
Now if the circle 7 is made sufficiently small, it follows from
the continuity of f(z) that \f(z) — f(a) \ can be made as small as
desired. On the other hand, it follows from (138-3) that the
value of the integral (140-4) is independent of the radius p of
the circle 7, so long as 7 is interior to R. Thus the left-hand
member of (140-5) is independent of p; and since M —*0 when
p —> 0, it follows that the value of the integral is zero.
Accordingly, (140-2) becomes
(140-6) -p^ = 2nf(a),
c/C ^ a
where a, which plays the role of a parameter, is any point interior
to C. Denote the variable of integration in (140-6) by f, and
let z be any point interior to C; then (140-6) can be written as
*» -
§140 COMPLEX VARIABLE 463
The relationship stated by (140-7) is known as Cauchy's integral
formula.
It is not difficult to show that an integral of the form (140-7)
can be differentiated with respect to the parameter z as many
times as desired, * so that
jw-—. - /(r)
(140-8) """' ~
In fact, if /(«) is any continuous (not necessarily analytic) function
of the complex variable z, then the integral
if?
defines an analytic function F(2). To show this, all that is
necessary is to form the difference quotient [F(z + Az) —
F(z)]/&z and to evaluate its limit as Az — > 0. It follows from
such a calculation that
The assertion made in Sec. 137, concerning the fact that the
continuity of the derivative of an analytic function follows from
the assumption that the derivative exists, is now made clear.
PROBLEMS
1. If
**
where C is the circle of radius 2 about the origin, find the values of
2. Apply Cauchy's integral formula to Prob. 7, Sec. 139. Use the
integrand in the form given in Prob. 8.
* Form the difference quotient [f(z + &z) — /(2)]/Az, and investigate the
behavior of the quotient as Az — » 0.
464 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §141
3. Evaluate with the aid of Cauchy's integral formula
where C is the circle |f | = 2.
4. What is the value of the integral of Prob. 3 when evaluated over
the circle |f — 1| = 1?
5. Evaluate
f- 2z - 1 J
— *,
Sc
where C is the circle \z\ = 1.
141. Taylor's Expansion. Let/(z)
be analytic in some region R, and
let C be a circle lying wholly in R
and having its center at a. If z is
any point interior to C (Fig. 134), then it follows from Cauchy's
integral formula that
FIQ. 134.
(141-1)
/(*)
= _! f
2*i JC f -
.
But
and substi-
1 - «• "'"'*' ' " ' 1 - «'
tuting this expression, with i = (z — a)/(f — a), in (141-1)
leads to
where
tori
Making use of (140-8) gives
(141-2) /(Z)=/(a)+/'(a)(2-a)
(f-a)-(r -«)">•
f"(a) , .,
-77T (2 - °)
§142 COMPLEX VARIABLE 465
By taking n sufficiently large, the modulus of Rn may be made
as small as desired. In order to show this fact, let the maximum
value attained by the modulus of /(£) on C be M, the radius of
the circle C be r, and the modulus of z — a be p. Then |f — z\
> r — p, and
r
Je
e r ~
p^ M27rr = Mr /pV
27r r»(r - p) " r - p \r/
Since p/r < 1, it follows that lim \Rn\ = 0 for every z interior
n — > oo
to C.
Thus, one can write the infinite series
/(z) = /(a) +/'(a)(* - a) + (2 - o)» + • • •
which converges to f(z) at every point z interior to any circle
C that lies entirely within the region R in which f(z) is analytic.
This series is known as the Taylor's series.*
PROBLEMS
1. Obtain the Taylor's series expansions, about 2 = 0, for the follow-
ing functions :
(a) es, (b) sin z, (c) cos z, (d) log (1 + i).
2. Verify the following expansions:
(a) tan z = z + -» + -r=+ • • • ;
(b) sinh 3 = * + fj + f]+''';
(c) cosh z = l+oi + T-f+*-';
142. Conformal Mapping. It was mentioned in Sec. 135 that
the functional relationship w = f(z) sets up a correspondence
* For a more extensive treatment, see D. R. Curtiss, Analytic Functions
of a Complex Variable; E. J. Townsend, Functions of a Complex Variable,
H. Burkhardt and S. E. Rasor, Theory of Functions of a Complex Variable;
466 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §142
between the points z = x + iy, of the complex z-plane, and
w = u + iv, of the complex w-plane. If w = /(z) is analytic in
some region R of the 2-plane, then the totality of values w belongs
to some region Rf of the w-plane, and it is said that the region R
maps into the region R1. If C is some curve drawn in the region
R and the point z is allowed to move along C, then the corre-
sponding point w will trace a curve C' in the w-plane (Fig. 135),
and C" is called the map of the curve C.
The relationship of the curves C and C' is interesting. Con-
sider a pair of points z and z + &z on C, and let the arc length
between them be As. = PQ. The corresponding points in the
w+Aw
w
FIG. 135.
region R' are denoted by w and w + Aw, and the arc length
between them by As7 = P'Q' . Since the ratio of the arc lengths
has the same limit as the ratio of the lengths of the corresponding
chords,
Aw
= lim
= lim
lim A _ A1AXA | A | AXA-l*. .
The function w = f(z) is assumed to be analytic, so that dw/dz
has a unique value regardless of the manner in which Az — > 0.
Hence, the transformation causes elements of arc, passing through
P in any direction, to experience a change in length whose
magnitude is given by the value of the modulus of dw/dz at P.
For example, if w = z3, then the linear dimensions at the point
3=1 are stretched threefold, but at the point z = 1 + i they
are multiplied by 6.
It will be shown next that the argument of dw/dz determines
the orientation of the element of arc As' relative to As. The
argument of the complex number Az is measured by the angle 9
made by the chord PQ with the #-axis, while arg Aw measures
§143
COMPLEX VARIABLE
467
the corresponding angle 0' between the w-axis and the chord
P'Q'. -Hence, the difference between the angles 0' and 0 is
equal to
A * Aw
arg Aw — arg Az = arg — >
for the difference of the arguments of two complex numbers is
equal to the argument of their quotient. As A£ — » 0, the vectors
Az and Aw tend to coincidence with the tangents to C at P and
C" at P', respectively, and hence* arg dw/dz is the angle of
rotation of the element of arc As' relative to As. It follows
immediately from this statement that if Ci and C^ are two curves
which intersect at P at an angle r (Fig. 136), then the corre-
Fio. 136.
spending curves C[ and C£ in the w-planc also intersect at an
angle r, for the tangents to these curves are rotated through the
same angle.
A transformation that preserves angles is called con/ormaZ, and
thus one can state the following theorem:
THEOREM. The mapping performed by an analytic function
f(z) is conformal at all points of the z-plane where f'(z) 7* 0.
143. Method of Conjugate Functions. The angle-preserving
property of the transformations by analytic functions has many
immediate and important physical applications.
For example, if an incompressible fluid flows over a plane
with a velocity potential 3>(x, y) (so that vx = d$/dx,vy — d$/dy},
then it is known that the stream lines will be directed at right
angles to the equipotential curves <£(x, y} = const. Moreover,
it was shown f that the functions $ and ^f satisfy the Cauchy-
* Note that this statement assumes that dw/dz 9* 0 at the point P.
t See Sec. 66.
468 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §143
Riemann equations, and hence one can assert that the functions
<£ and ^ are the real and imaginary parts, respectively, of some
analytic f unction /(z), that is,
/(*) - <*>(*, y) + i*(x, y).
Now, let w = f(z) = $ + i^, and consider the two families
of curves in the w-plane defined by
(143-1) $(#, y} — const. and
(x, y) = const.
The orthogonality of the curves <i> = const, and ^ = const, in
the z-plane follows at once from the conf ormal properties of the
transformation by the analytic function /(z). For <f> = const.
and ^ = const, represent a net of orthogonal lines (Fig. 137)
$= const
- const
-X
FIG. 137.
parallel to the coordinate axes in the w-plane, and they are
transformed by the analytic function w = $ + i^ into a net of
orthogonal curves in the z-plane.
It is obvious then that every analytic f unction /(z) = u(x, y) +
iv(x, y) furnishes a pair of real functions of the variables
x and y, namely, u(x, y) and v(x, y), each of which is a solution
of Laplace's equation. The functions u(x, y) and v(x, y) are
called conjugate functions, and the method of obtaining solutions
of Laplace's equation with the aid of analytic functions of a
complex variable is called the method of conjugate functions.
Example. The process of obtaining pairs of conjugate functions from
analytic functions is indicated in the following example. Let
then,
w - u + iv = sin z = sin (x + iy) ;
u + iv = sin x cos iy + cos x sin iy,
= sin x cosh y + i cos x sinh ?/,
§143 COMPLEX VARIABLE 469
so that
u(x, y) = sin x cosh y,
v(x> y) = cos x sinh y.
It is not difficult to show that the inverse of an analytic func-
tion is, in general, analytic. Thus, the solution of the equations
u = &(x, y) and v = W(x, y)
for x and y in terms of u and v furnishes one with a pair of
functions
x = <p(u, v) and y = \l/(u, v)
that satisfy Laplace's equation in which u and v are the inde-
pendent variables.
The following three sections are devoted to an exposition of
the method of conjugate functions as it is employed in solving
important engineering problems *
PROBLEMS
1. Discuss the mapping properties of the transformations defined
by the following functions. Draw the families of curves u = const,
and v — const.
(a) w = u-\-iv = z-\-a, where a is a constant;
(b) w = bz, where b is a constant;
(c) w = bz + a, where a and b are constants;
(d) w = z2;
(e) w = l/z.
2. Obtain pairs of conjugate functions from
(a) w = cos z-,
(b) w = e*;
(c) w = z3;
(d) w = log z;
(e) w = l/z.
* The material contained in Sees. 144 to 146 is extracted from a lecture on
conformal representation, which was delivered by invitation at the S. P. E. E.
Summer Session for Teachers of Mathematics to Engineering Students at
Minneapolis, in September, 1931, by Dr. Warren Weaver, director of the
Division of Natural Scien es of the Rockefeller Foundation, and formerly
professor of mathematics at the University of Wisconsin.
The authors did not feel that they could improve upon the lucidity and
clarity of Dr. Weaver's exposition of the subject and are grateful for his kind
permission to make use of the lecture, which was printed in the October,
1932, issue of the American Mathematical Monthly.
470 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §144
144. Problems Solvable by Conjugate Functions. Specific
examples of the method of conjugate functions will be given later,
but it may be well to indicate here two general sorts of problems.
Suppose that an analytic function w = u + iv = f(z) = f(x + ly)
maps a curve C of the 2-plane (see Fig. 138), whose equation
w-plom*
z-plome
FIG. 138.
is y = <p(x), onto the entire real axis v = 0 of the w-plane.
will obviously occur if, and only if,
This
v[x,
s 0.
Then the function
(z, y) s= v(x, y)
clearly is a solution of Laplace's equation that reduces to zero
on the curve C. In an important class of problems of applied
mathematics, one requires a solution of Laplace's equation that
reduces to zero, or some other constant, on some given curve.
Thus, one may, so to speak, go at such problems backward;
and, by plotting in the z-plane the curves u(x, y) = const, and
v(x, y) = const., he finds for what curves C a given analytic
function solves the above problem. Similarly, one may inter-
change the roles of u, v and x, y and may plot in the w-plane the
curves x(u, v) = const, and y(u, v) = const. Thus a properly
drawn picture of the plane transformation indicates to the
eye what problems, of this sort, are solved by a given analytic
function. It must be emphasized that the picture must be
" properly drawn"; that is, one requires, in one plane, the
§146 COMPLEX VARIABLE 471
two families of curves obtained by setting equal to various con-
stants the coordinate variables of the other plane.
In a second and more general sort of problem, it is necessary
to obtain a solution <£(#, y) of Laplace's equation which, on a
given curve C whose equation is y — <p(x), reduces to some given
function <£>*(#, y}. The previous problem is clearly a very special
case of this second problem. Suppose, now, that an analytic
function w = f(z) map the curve C of the 2-plane onto the axis
of reals v = 0, of the w-plane. Since the curve C maps onto
v = 0 in the w-plane, v[x, <p(x)] = 0, and the values of <£* at
points on C are equal to the values of
**l*(u, 0), y(u, 0)] sfc.dO
at the corresponding points on the transformed curve v — 0.
Suppose now that the function ^(w, v) be a solution of Laplace's
equation (u and v being viewed as independent variables), such
that
y(u, 0) ss $>*(u).
It is easily checked that
&(x, y} = V[u(x, y), v(x, y}}
is a solution of Laplace's equation, x and y being viewed as
independent variables. Moreover,' on the curve C one has
*[*, v(x)] = *(w, 0) = **(u) = **(*, y),
so that $ is the solution sought.
The chief service, in this case, of the method of conjugate
functions, is that the form of the boundary condition is much
simplified. Rather than seeking a function that takes on
prescribed values on some curve C, one has rather to find a func-
tion that takes on prescribed values on a straight line, namely,
the axis of abscissas. This latter problem is so much simpler
than the former that it can, indeed, be solved in general form for
a very general function <!>„,. This solution will be referred to
later, in Sec. 146c.
145. Examples of Conformal Maps. As a preparation for the
consideration of applications, this section will present six specific
instances of the conformal mapping of one plane on another.
The examples chosen are not precisely those which one would
select if, building up from the simplest cases, one were to study
472 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §145
the mathematical theory in detail. The examples are chosen
for their characteristic features and because of their important
and direct applications. The first case is:
a. The Transformation w = zmj m a Positive Integer. If one
write both z and w in polar form, so that
then
and
z = re1*,
w = Re1*,
w == Re1* = zm = rmeim<p,
R = r™,
<$> = m<p.
Thus the curve r = const, in the 2-plane (that is, a circle about
the origin) transforms into a curve R = const, in the w-plane
FIG. 139.
(also a circle about the origin), the radius of the circle in the
w-plane being equal to the rath power of the radius of the circle
in the 2-plane. Also, a radial line <p = const, in the z-plane
transforms into a new radial line 3> = const., the amplitude angle
for the transformed radial line being ra times the amplitude
angle of the original radial line. Thus, a sector of the 2-plane
of central angle 27r/ra is "fanned out" to cover the entire w-plane,
this sector also being stretched radially (see Fig. 139, drawn for
ra = 3). One notes the characteristic feature that a set of
orthogonal curves in one plane transform into a set of orthogonal
curves in the other plane.
§146 COMPLEX VARIABLE 473
This example suggests several interesting questions which
cannot be discussed here. The " angle-true " property clearly
does not hold at the origin, which indicates that this point
deserves special study. Further, it is clear that only a portion
of the 2-plane maps onto the entire w-plane. In the case for
which the figure is drawn, it would require three w-planes, so to
speak, if the entire 2-plane were to be unambiguously mapped.
This consideration leads to the use of many-sheeted surfaces,
called Riemann surfaces. Such questions and apparent diffi-
culties correctly indicate that a thorough knowledge of the mathe-
matical theory of analytical functions is essential to a proper
and complete understanding of even simple instances of conformal
representation.1
To get a clear idea of the way in which the 2-plane maps onto
the w-plane, one may choose various convenient families of curves
in one plane and determine the corresponding curves in the other
plane. The resulting picture, as was mentioned earlier, does not
give any indication of the immediate physical applications of the
transformation in question unless one of the sets of curves, in
one plane or the other, consists of the straight lines parallel to the
coordinate axes. It should thus be clear that Fig. 139 does not
give a direct indication of the type of problem immediately
solvable by the transformation w = z3. The curves in the
w-plane obtained by setting x — const, and y = const, are, in
fact, cubic curves; and no simple physical problem is directly
solved by this transformation. This transformation may, how-
ever, be used to solve various physical problems for a wedge-
shaped region, since the bounding curve C of such a wedge
(say the line <p = 0 and the line <p = 7r/3) is transformed into
a curve C' of the w-plane that consists of the entire real axis.
Thus the transformation can be used, in the way indicated in
Sec. 144, to solve problems in which one desires a solution of
Laplace's equation that reduces to a given function (or a con-
stant) on the boundary of a wedge.
H)
6. The Transformation w = - — ~ — ~* This again is a trans-
1 BIEBERBACH, L., Einfuhrung in die konforme Abbildung, Berlin, 1927;
LEWENT, L., Conformal Representation, London, 1925; OSGOOD, W. F.,
Lehrbuch der Funktionentheorie, vol. 1, Chap. XIV.
474 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §146
formation that does not have immediate applicability. It has,
however, interesting features, and subsequent discussion will
indicate how it may be made to serve a practical purpose.
FIG. 140.
If, as before, one write z in the polar form re1*, then
w = u + iv =
z)
"27
so that
1 ( JL l\ _L ' ! ( l\ '
= 2 V V C°S * + Z 2 V ~ V Sm ^
I/ i A
W = 2 V r/ C°S 9|
Thus, ^ and r being eliminated in turn,
cos2 <p sin
2
2 ~ 4'
= 1.
From these equations, it follows by inspection that the circles
r = const, of the 2-plane transform into a family of ellipses of
the w-plane (see Fig. 140), the ellipses being confocal, since
§146 COMPLEX VARIABLE 475
V + r) ~ V ~~ r) " 4 =
It is also clear that two circles of reciprocal radii transform into
the same ellipse. Similarly, the radial lines <p = const, of the
z-plane transform into a family of hyperbolas which, again, arc
confocal, since
cos2 <f> + sin2 <? = 1 = const.
Thus the exterior of the unit circle of the z-plane transforms
into the entire w-plane. The unit circle itself "flattens out"
to form the segment from — 1 to + 1 of the real axis of the w-plane.
All larger circles are less strenuously "flattened out" and form
ellipses, while the radial lines of the z-plane form the associated
confocal hyperbolas of the w-plane. A similar statement can be
made for the inside of the unit circle.
c. The Transformation w = ez. If one set w ss Re1* and
z = x + iy, then
Re1* = e*+*v = ex - elv,
so that
7? = ex,
$ = y.
It is thus clear that vertical lines of the z-plane map into circles
of the w-plane, the radius being greater or less than 1, accord-
ing as # is positive or negative. Horizontal lines of the z-plane,
on the other hand, map into the radial lines of the w-plane, and
it is clear that any horizontal strip of the z-plane of height 2ir
will cover the entire w-plane once (see Fig. 141).
The curves in the w-plane of Fig. 141 are drawn by setting
equal to a constant one or the other of the coordinates of the
z-plane. Thus these curves give direct indication of physical
problems to which this analytic function may be applied. For
example, one could obtain the electrostatic field due to a charged
right circular cylinder, the lines of flow from a single line source
of current or liquid, the circulation of a liquid around a cylindrical
obstacle, etc.
By considering this example in conjunction with the preceding
example, one gives new significance to Fig. 140. In fact, if one
starts with the z-plane of Fig. 141 and then uses the w-plane of
Fig. 141 as the z-plane of Fig. 140, it is clear that the curves
drawn in the w-plane of Fig. 140 then are obtainable by setting
476 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §145
equal to various constants the coordinates of the 2-plane of
Fig. 141. That is to say, the w-plane curves of Fig. 140 give
direct evidence of physical problems that can be solved by the
pair of transformations
d. The Transformation w = cosh z. If, in the two preceding
equations, one eliminates the intermediate variable z\ (so he may
' _J_L_ I L-
w-plane
Z-plane
Fia. 141
pass directly from the 2-plane of Fig. 141 to the w-plane of Fig.
140), the result is
J_ e~ z
w =
= cosh z.
Thus
u -J- iv = cosh (x + iy) = cosh x cosh iy + sinh x sinh iyf
= cosh x cos y + i sinh x sin y,
so that
u = cosh x cos y
v = sinh x sin y,
or
^.2 ..2
= 1,
cosh2 # sinh2 x
u2 v*
cos2
sin2 y
= 1.
§145
COMPLEX VARIABLE
477
This transformation is shown in Fig. 142, and it may be used to
obtain the electrostatic field due to an elliptic cylinder, the
electrostatic field due to a charged plane from which a strip has
been removed, the circulation of liquid around an elliptical
cylinder, the flow of liquid through a slit in a plane, etc.
The transformation from the z-plane to the w-plane may be
described geometrically as follows: Consider the horizontal strip
of the 2-plane between the lines y = 0 and y = TT; and think
of these lines as being broken and pivoted at the points where
x = 0. Rotate the strip 90° counterclockwise, and at the same
2 -plane
FIG. 142.
time fold each of the broken lines y = 0 and y = ir back on itself,
the strip thus being doubly "fanned out" so as to cover the
entire w-plane.
e. The Transformation w = z + ez. One has
so that
u + iv = x + iy + ex+lv,
= x + iy + ex(cos y + i sin y),
u = x + ex cos y,
v = y + ex sin y.
This transformation is shown in Fig. 143. If one considers the
portion of the z-plane between the lines y = ±TT, then the portion
of the strip to the right of x = — 1 is to be "fanned out" by
rotating the portion of y = +1 (to the right of x = —1) counter-
clockwise and the portion of y = — 1 (to the right of x = — 1)
clockwise until each line is folded back on itself. This trans-
478 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §145
formation gives the electrostatic field at the edge of a parallel
plate condenser, the flow of liquid out of a channel into an open
sea, etc.
/. The Schwartz Transformation. The transformations just
considered are simple examples and are necessarily very special
in character. . This list of illustrations will be concluded by a
FIG. 143.
*Z *3
x-p1o»ne
*4 «5
w-plome
Fio. 144
more general transformation. Suppose one has (see Fig. 144) a
rectilinear polygon, in the w-plane, whose sides change direction
by an angle aw when one passes the ith vertex, going around the
boundary of the polygon so that the interior lies to the left.
The interior of this polygon can be mapped onto the upper
half 3-plane by the transformation
w
dz
<«-
§146 COMPLEX VARIABLE 479
where «i, s*, • ' • , Sn are the (real) points, on the #-axis of the
2-plane, onto which map the first, second, • • • , nth vertex
of the polygon, and where A and B are constants which are to be
determined to fit the scale and location of the polygon. Three
of the points zl may be chosen at will, and the values of the
remaining ones may be calculated.
This theorem may be used to find, for example, the analytic
transformation that solves the problem of determining the
electrostatic field around a charged cylindrical conductor of any
polygonal cross section. It should be noted, however, that one
requires for this purpose the function y(u, v), whereas the
theorem gives one w as a function of z. It is often exceedingly
difficult and laborious to solve this relation for z as a function of
w, so that one may obtain the function y. It should further be
remarked that this theorem may be applied to polygons some
of whose vertices are not located in the finite plane and that
the theorem is of wide applicability and importance in connec-
tions less direct and simple than the one just mentioned.
146. Applications of Conformal Representation, a. Applica-
tions to Cartography. It is natural that a mathematical theory
which discusses the "mapping'' of one plane on another should
have application to the problems connected with the drawing of
geographic maps. Since the surface of a sphere cannot be made
plane without distortion of some sort, one has to decide, when
mapping a portion of the sphere on a plane, what type of distor-
tion to choose and what to avoid. For some purposes, it is
essential that areas be represented properly; for other purposes,
it is most important that the angles on the map faithfully repre-
sent the actual angles on the sphere.
The first problem, in conveniently mapping a sphere on a plane,
is to map the sphere on the plane in some fashion or other and
then, if this fashion be unsatisfactory, to remap this plane onto a
second plane. The first problem can be done in a wide variety
of ways1 which include, as important examples, stereographic
projection and Mercator's projection. Both these examples
are conformal projections, in that they preserve the true values
of all angles. Having once mapped the sphere on the plane
(or on a portion of the plane), one may now remap onto a second
1 The Encyclopaedia Britannica article on maps lists and discusses nearly
thirty such projections actually used in map making.
480 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §146
plane, and it is here that the theory of conformal representation
finds its application ; for one can determine the analytic function
that will conformally remap the original map onto a new region
of any desired shape and size. Not only are all angles preserved
in this process of conformal remapping, but the distortion in the
neighborhood of a point is always a pure magnification. Thus
the shapes of all small objects or regions are preserved. Such
maps do not give a true representation of areas, and for this rea-
son many maps are based on compromises between conformal
transformations and area-preserving transformations.
b. Applications to Hydrodynamics. When the velocities of all
particles of a moving liquid lie in planes parallel to one plane
that we may conveniently choose as the xy-p\&ue and when
all particles having the same x and y have equal velocities, then
the motion is said to be two-dimensional. Such cases clearly
arise if a very thin sheet of liquid is flowing in some manner over
a plane or if a thick layer of liquid circulates over a plane, there
being no motion and no variation of motion normal to the plane.
Let the x- and ^-components of velocity at any point (x, y) be u
and y, respectively. The motion is said to be irrotational if the
curl of the velocity vector vanishes. Analytically, this demands
that
— — —
dy ~~ dx
whereas physically it states that the angular velocity of an
infinitesimal portion of the liquid is zero. The equation just
written assures that
— (u dx + v dy)
is the perfect differential of some function, say 4>. This function
is known as the velocity potential, since by a comparison of the
two equations
d$ = — u dx — v dy,
— dx + — dy,
dx dy yy
it follows that
/m*i\
(146-1)
Now, if the liquid be incompressible, the amount of it that
flows into any volume in a , given time must equal the amount
§146 COMPLEX VARIABLE 481
that flows out. This demand imposes on the components of
velocity the restriction that
dx dy '
this being known as the equation of continuity. From the last
two equations, it follows that
d2<l> d2<l>
f.
d?/
_ 4. — = v2$ = 0
r\ 9 I f. n - v ^* ^^ V/.
2 2
That is, the velocity potential satisfies Laplace's equation.
Just as the vanishing of the curl of the velocity demands that
u dx + v dy be an exact differential, so the equation of continuity
demands that v dx — u dy be an exact differential of some func-
tion, say ty. That is,
d^f = v dx — u dy,
d* , , 3V j
d* = -dx + -dy,
so that
(146-2) „ = «* u = - £.
6x dy
From (146-1) and (146-2), it follows at once that
d$d* ,d$<M
dx dx + dy dy 7
which expresses the geometric fact that the curves $ = const.
and ^ = const, intersect everywhere orthogonally. It is clear
from (146-1) that there is no component of velocity in the
direction of the curves on which <l> is a constant, so that the veloc-
ity of the liquid is everywhere orthogonal to the equipotential
curves $ = const. That is, the curves ^ = const, depict
everywhere the direction of flow. For this reason, ^ is called
the stream function and the curves ^ = const, are called the
stream lines. From (146-2) and the vanishing of the curl of
the velocity, it follows that the stream function ^ is also a solu-
tion of Laplace's equation.
Thus, the velocity potential $ and the stream function ^
in the case of the irrotational flow of a perfect incompressible
liquid both satisfy Laplace's equation, and the curves & = const.
482 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §146
and ^ = const, form two orthogonal families. Every analytic
function therefore furnishes the solution to four such problems,
the four solutions resulting from the fact that one may choose
the pair x, y or the pair u, v as independent variables, and that
one may interchange the roles of the potential function and the
stream function. Figure 142, for example, indicates two of
the four problems solved by the analytic transformation w =
cosh z. If one treats u and v as the independent variables and
identifies the (solid) curves y(u, v) = const, in the w-plane
with the curves <f> = const., then the dotted curves x(u, v) =
ty = const, give the stream lines, and one has solved the prob-
lem of the circulation of liquid around an elliptic cylinder. If,
however, one sets y(u, v) = & and x(u, v) = <£, then the solid
curves of the w-plane are the stream lines, and one has solved
the problem of the flow of liquid through a slit. The other two
problems solved by this same function are to be obtained by
drawing, in the 2-plane, the curves u(x, y) = const, and v(x, y) =
const, and identifying ^ and $ with u and t;, and vice versa.
The 2-plane curves u = const, and v = const, are very com-
plicated and do not correspond to any simple or important
physical problem, and hence they are not drawn on the figure.
In fact, it is usually the case that only two of the possible four
problems are sufficiently simple to be of any practical use.
It should be emphasized that it is never sufficient, in obtaining
the analytical solution of a definite physical problem, merely
to know that certain functions satisfy Laplace's equation.
One must also have certain boundary conditions. The graphs
shown above disclose to the eye what physical problem has been
solved precisely because they show what sort of boundary condi-
tions are satisfied. For example, if the dotted curves of Fig. 142
are stream lines, then the problem solved is the circulation around
an elliptical obstacle just because these dotted stream lines
satisfy the boundary condition for such a problem; namely,
because the flow at any point on the boundary of the obstacle is
parallel to the boundary of the obstacle.
It is interesting to note that this same transformation w =
cosh z (or, slightly more generally, w = a cosh z) can be used to
solve a hydrodynamic problem of a different sort. When liquid
seeps through a porous soil, it is found that the component in
any direction of the velocity of the liquid is proportional to the
§146
COMPLEX VARIABLE
483
negative pressure gradient in that same direction. Thus, in a
problem of two-dimensional flow,
If these values be inserted in the equation of continuity, namely,
in the equation
the result is
0.
Suppose, then, one considers the problem of the seepage flow
under a gravity dam which rests on material that permits such
seepage. One seeks (see Fig. 145) a function p that satisfies
624/p0 -=-_=
P'Po
FIG. 145.
Laplace's equation and that satisfies certain boundary conditions
on the surface of the ground. That is, the pressure must be
uniform on the surface of the ground upstream from the heel of
the dam and zero on the surface of the ground downstream
from the toe of the dam. If we choose a system of cartesian
coordinates u, v with origin at the midpoint of the base of the
dam (Fig. 145) and w-axis on the surface of the ground, then
it is easily checked that p(u, v) = pvy(u, v)/*, where
w = u + iv = a cosh (x + iy),
484 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §146
satisfies the demands of the problem. In fact, it was seen in
Sec. 145d, where the transformation w = cosh z was studied,
that the line y = TT of the 2-plane folds up to produce the portion
to the left of u = — 1 of the ^-axis in the w-plane, and the line
y = 0 of the z-plane folds up to produce the portion to the right
of u = +1 of the u-axis. The introduction of the factor a
in the transformation merely makes the width of the base of the
dam 2a rather than 2. These remarks show that p(u, v) reduces
to the constant TT on the surface of the ground upstream from the
heel of the dam. If the head above the dam is such as to produce
a hydrostatic pressure po, one merely has to set
P(U, v) =
71
One may now easily find the distribution of uplift pressure across
the base of the dam. In fact, the base of the dam is the repre-
sentation, in the i^-plane, of the line x = 0, 0 ^ y ^ TT, of the
:n/-plane. Hence, on the base of the dam the equations
u = a cosh x cos y,
v = a sinh x sin y
reduce to
u = a cos y,
v = 0,
so that
p(u, 0) = — cos"1 —
TT a
This curve is drawn in the figure. The total uplift pressure
(per foot of dam)
/•+a
PO I
= — I
TT J-a
17
cos"1 - du =
which is what the uplift pressure would be if the entire base of
the dam were subjected to a head just one-half the head above the
dam or if the pressure decreased uniformly (linearly) from the
static head pQ at the heel to the value zero at the toe. The point
of application of the resultant uplift is easily calculated to be
at a' distance b = 3a/4 from the heel of the dam.
c. Applications to Elasticity. If opposing couples be applied
to the ends of a right cylinder or prism of homogeneous material,
§146 COMPLEX VARIABLE 485
the cylinder twists and shearing stresses are developed. Choose
the axis of the prism for the z-axis of a rectangular system of
coordinates. The angle of twist per unit length, say r, and the
shearing stresses, due to an applied couple T, can both be cal-
culated if one can determine a function <£(#, y) satisfying Laplace's
equation and reducing, on the boundary of a section of the prism,
to the function 3>* = (x* + y*)/2. In fact,1
where
C = 2(7 JJ ($ - $*) do: dy,
in which (? is the modulus of rigidity of the material, whereas the
shearing stresses are given by
X, = G
1
/ \
\ /
Exact analytical solutions of
this important technical problem
i. u i_x • j r i FIG. 146.
nave been obtained ior several
simple sections, notably circular, elliptical, rectangular, and tri-
angular.2 Only recently3 the problem was solved for an infinite
T section (see Fig. 146). From the general discussion given in
Sec. 144, it is clear that, to solve this latter problem, one requires
first an analytic function that will map the boundary of this T
section onto the entire real axis of the new w-plane. This sec-
tion, moreover, is a rectilinear polygon, so that one can use the
Schwartz transformation theory to produce the desired analytic
relation. One finds that the desired mapping is carried out by
the function
1 LOVE, A. E. H., Theory of Elasticity, 3d ed., pp. 315-333, 1920.
2 TRAYER, W., and H. W. MARCH, The Torsion of Members Having Sec-
tions Common in Aircraft Construction, Bur. Aeronautics Navy Dept.,
Separate Rept. 334; also contained in Nat. Adv. Comm. Aeronautics, loth
Ann. Rept., 1929, pp. 675-719.
3 SOKOLNIKOFF, I. S., On a Solution of Laplace's Equation with an Appli-
cation to the Torsion Problem for a Polygon with Reentrant Angles, Trans.
Amer. Math. Soc., vol. 33, pp. 719-732.
486 MATHEMATICS FOR ENGINEERS AND PHYSICISTS ?U6
A C (W2 - 1)*S i_ D
Z = A I 7 — 9 «v- O^ -H #,
J (^2 - a2)
d
. x .
= - log (u> +
where the first line is furnished directly by the Schwartz theorem
and where, in the second line, the constants a, A, B have been
evaluated so as to fit the dimensions and location of the T
section.
It is next necessary to break z up into its real and imaginary parts
so as to obtain x and y as functions of u and v. These values,
when substituted into
»
give, because v = 0 on the boundary of the section, the function
**[*(u, 0), y(u, 0)] s *«,(*).
The remaining essential step is to obtain a function ^(u, v)
satisfying Laplace's equation and reducing, on the axis of
reals v = 0, to the function $#(u). Such a function is1
TT -oo P2 - 2£pcos0+ {*'
where
w = u + z'y = pe*e.
The solution to the original problem is then given, as was earlier
indicated in S"ec. 144, by <£ = ^. It is obviously a difficult and
laborious job to carry out these calculations, but formulas have
been obtained, in the paper referred to, from which practical
calculations can be made.
d. Applications to Electrostatics. The methods of complex
variable theory are peculiarly applicable to two-dimensional
electrical problems. In order that the problems be two-dimen-
sional, we shall understand that the conductors under considera-
tion are exceedingly long cylinders whose axes are normal to the
z = x + iy plane. Under these circumstances the various
1 SOKOLNIKOFF, I. S., On a Solution of Laplace's Equation with an Appli-
cation to the Torsion Problem for a Polygon with Reentrant Angles, Trans.
Amer. Math. Soc., vol. 33, pp. 71&-732. This formula is the general solution,
spoken of in Sec. 144, of Laplace's equation subject to specified boundary
values on the entire axis of abscissas.
§146
COMPLEX VARIABLE
487
electrical quantities do not change appreciably in the direction
normal to the 2-plane, and one has to determine these quantities
as functions of x and y only. In certain problems, one or more
of the conductors present will have very small cross sections and
will be given a charge of, say ef per unit length. Such a con-
ductor will be called a line charge of strength ef.
The electrostatic problem for such conductors is solved when
one has obtained a function $(x, ?/), known as the electrostatic
potential, satisfying the following conditions:1
(a)
V2* = ^-^ 4- --— = 0
d*2 + dw2
(146-3)
at all points in free space.
(6) <1> reduces, on the surface of the kth conductor,
to a constant 3>fc.
(c) In the neighborhood of a line charge of strength
ef, $ becomes infinite as
— e' log r
2^ '
where r measures distance to the line.
(d) $ behaves at infinity as
log R2e'
27~~'
where Se' is the total charge per unit length of all
conductors present and where R measures distance
from some reference point in the finite plane. In
case Se' = 0, $ approaches zero as 1/R.
It is readily shown, by standard methods, that the solution of
such a problem is unique. This remark is of great practical
importance, since it assures one that a function <£ satisfying these
conditions is, however it may have been obtained, the correct
solution of the physical problem.
Physically one wishes to know the distribution of charge and
the electrostatic force at any point. These data may be obtained
1 See MASON, M., and W. WEAVER, The Electromagnetic Field, pp. 134, 146,
1929; and REIMANN- WEBER, Die Differentialgleichungen der Physik, vol. 2, p.
290, 1927. The units used in the above discussion are the rational units
used in Mason and Weaver, loc. cit.
488 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §146
from the function $ in the following manner: The component Ea
in any direction s of the electrostatic force per unit charge is
given in terms of <£ by the relation
and the surface density of charge 77 on any conductor is given by
77 = — )
dn
where n measures distance along the external normal to the
conductor in question.
Now if
w = u + iv = f(z) = f(x + iy)
and if the function
$(.?, y) = u(x, y)
satisfies condition (146-3), then
jP _ d<£ du _ dv
the last step following from the Cauchy-Riemann equations
(135-6). Similarly
v dy dy dx
Thus,
the last step resulting from the fundamental fact that the value
of the derivative of an analytic function w is independent of the
mode in which z approaches zero. Now the complex number
a — ib is called the " conjugate" of the complex number a + ib
and one often denotes a conjugate by a bar, thus:
a — ib = a + bi.
With this standard notation, the "complex electric force"
E SE Ex + iEy is given by
(146-4) E m E, + iEy = - §?,
§146 COMPLEX VARIABLE 489
and the magnitude \fEl + El of the electrostatic force at any
point is given by
*=~
If one chooses $ s v(x, y), then (146-4) and (146-5) are
replaced by
(146-6) E = Ex + iEy = -i ^>
(146-7) VW+ El =
dw
Tz
Three types of electrostatic problems will now be briefly
considered. The first and simplest two-dimensional electrostatic
problem is that of a single long cylindrical conductor with a given
charge. One then seeks a function that satisfies Laplace's
equation and, in accordance with (146-3)6, reduces to a constant
on the curve that bounds a section of the conductor. This is
the analytical problem whose solution was indicated in Sec. 144.
One requires a function w = u + iv = f(z) = f(x + iy) such
that either a vertical straight line u = const, or a horizontal
straight line v = const, of the w-plane maps into the bounding
curve C of the conductor's section in the 2-plane. Then <£(z, y)
= u(x, y) or $(x, y) = v(x, y) solves the problem, and the
physically important quantities are given by (146-4), (146-5)
or by (146-6), (146-7), respectively.
Secondly, suppose that a single long cylindrical conductor is
in the presence of a parallel line charge of strength ef '. We
suppose the line charge to be outside the conductor. Let C
be the bounding curve in the 2-plane of a section of the conduc-
tor, and let the line charge be located at z = ZQ. We may
conveniently suppose the cylindrical conductor to be grounded,
so that we seek a solution of Laplace's equation that reduces
to zero on C and becomes infinite as indicated in (J46-3)c at
z = z0. Let f = f(z) transform C onto the entire axis of reals
and the exterior of C conformally upon the upper half f -plane.
Then if
the function 3> ss u(x, y) is the solution sought. In fact, for
values of z sufficiently close to 20, f(z) — f(zo) behaves, except for
490 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §146
a constant factor, as (z — ZQ). Thus, if one writes
then, for values of z very near to 20,
e' 1 e' 1 e'0
10 = 14 -|- iv = — — log -r -f- A = pr— log — — i p: — ~\~ A,
2?r re ZTT r &TT
where A remains finite as z = ZQ. Therefore,
where B remains finite as z = ZQ. Thus u(xy y) has the proper
type of infinity at z = 20. Furthermore, for points z on C,
/(«) is on the axis of reals in the f -plane, so that the modulus of
/(z) — fM equals the modulus of f(z) — /(ZQ). Hence the
modulus of
/(«>-/(*>)
is unity. However, since
log pe'*> = log p + zV,
it is clear that the real part of the logarithm of a complex quantity
is the logarithm of the modulus of the complex quantity. Since
the logarithm of unity is zero, it is clear that u vanishes on C.
As regards the behavior of u(x, y) at infinity, one notes that u
is the logarithm of the ratio of the (real) distances of f = f(z) to
f o = /(ZQ) and to f0 = /(ZQ). As z becomes infinite, this ratio
differs from unity by an amount whose leading term is equal to
or less than a constant times_the reciprocal distance from f(z)
to one of the points /(z0) or/(20). Thus the leading term in the
logarithm of this ratio is a constant times this reciprocal distance;
and $ = u behaves at °o in the required manner.
'In the third type of problem there are two conductors present,
one raised to the potential <f>o while the other is at a potential
zero. Thus, suppose that the cross section of two long cylindrical
conductors consists of two curves Co and Ci, such as those shown
in Fig. 147, which do not intersect at a finite point but which,
if one takes account of the intersection of BI and B0 and of AQ
and A i at z = <», divide the extended plane into two simply
connected regions, one of which may be called the "interior"
and the other the "exterior" of the closed curve C0 + Ci. Now
suppose that f = f(z} maps Co onto the entire negative axis of
§146
COMPLEX VARIABLE
491
reals in the f -plane, with the infinitely distant point along BQ
mapped onto f = 0, that f = f(z) also maps C\ onto the entire
positive axis of reals with the infinitely distant point along
BI mapped onto f = 0, and that f = /(z) maps the interior of
Co + Ci conformally on the upper half f -plane. Then, if
w = u + w = -° log/(z),
7T
the function $ = v(x, y) satisfies V2<l> = 0 at every point in the
interior of C0 + Ci, reduces to zero on Ci, and reduces to <f>0 on C0.
AO
4" plan
FIG. 147.
In fact, the imaginary part of the logarithm of a complex number
is merely the amplitude of the complex number; and for points
on Co, f(z) has an amplitude of TT, while for points on Ci, f(z) has
an amplitude of zero. Then,
ds
dw ___ $o dz
Hz ~~ 7/60'
and the electrostatic force is given by (146-5) and (146-7).
This third type of problem is of frequent and important
practical occurrence. Many electrical engineering problems
that have been solved by this method of conformal representa-
tion are referred to in an expository article, devoted largely to
the Schwarz transformation, by E. Weber.1 In an earlier article
in the same journal,2 for instance, the theory of conformal
representation is applied to the problem of studying the leakage
voltage and the breakdown potential between the high- and low-
potential portions of oil-immersed transformers. The cases
studied come under the third type of problem discussed above.
1 WEBER, E., Archiv fur Elektrotechnik, vol. 18, p. 174, 1926.
2 DREYFUS, L., Archiv fUr Elektrotechnik, vol. 13, p. 123, 1924.
CHAPTER XI
PROBABILITY
There is no branch of mathematics that is more intimately
connected with everyday experiences than the theory of proba-
bility. Recent developments in mathematical physics have
emphasized anew the great importance of this theory in every
branch of the physical sciences. This chapter sets forth the bare
outline of those fundamental facts of the theory of probability
which should form a part of the minimum equipment of every
student of science.
147. Fundamental Notions. Asking for the probability or
for a measure of the happening of any event implies the possi-
bility of the non-occurrence of this event. Unless there exists
some ignorance concerning the happening of an event, the
problem does not belong to the theory of probability. Thus, the
question "What is the probability that New Year's day in 1984
will fall on Monday?" is trivial, inasmuch as this question can
be settled by referring to a calendar. On the other hand, the
query "What is the probability of drawing the ace of hearts
from a deck of 52 cards?" constitutes a problem to which the
theory of probability gives a definite answer. In fact, one can
reason as follows: Granting that the deck is perfect, one card is
just as likely to appear as any other and, since there are 52 cards,
the chance that the ace of hearts will be drawn is 1 out of 52.
The words "just as likely," used in the preceding sentence, imply
the existence of the ignorance that is essential to remove any
problem 'of probability from triviality. The term "equally
likely," or "just as likely," applied to a future event that can
happen or fail to happen in a certain number of ways, indicates
that the possible ways are so related that there is no reason for
expecting the occurrence of any one of them rather than that of
any other.
If an event can happen in N ways, which are equally likely,
and if, among these N ways, m are favorable, then the probability
492
§147 PROBABILITY 493
of the occurrence of the event in a single trial is
m
P-y
Thus, the probability that the six will appear when a die is
thrown is ^, since the total number of ways in which a die can
fall is 6, and of these six ways only one is favorable. The proba-
bility of drawing a heart from a deck of 52 cards is J^, since there
are 13 hearts and the total number of equally likely ways in
which a card can be drawn is 52.
It is clear that, if an event is certain to happen, then the
probability of its occurrence is 1, for all the possible ways are
favorable. On the other hand, if an event is certain not to occur,
the probability of its occurrence is zero. It is clear also that, if
the probability of the happening of an event is p, then the proba-
bility of its failure to happen is
q = 1 - p.
The concept of " equally likely " plays a fundamental role in the
theory of probability. The need for caution and a careful
analysis of the problem will be illustrated by several examples.
Let two coins be tossed simultaneously. What will be the
probability that they both show heads? The following reasoning
is at fault. The total number of ways in which the coins can fall
is three, since the possible combinations are two heads, two tails,
and a head and a tail. Of these three ways, only one is favorable,
and therefore the probability is J^. The fault in this reasoning
lies in the failure to account for all the equally likely cases.
The number of ways in which one head and one tail can fall is
two, since the head can appear on the first coin and the tail on
the second, or the head can appear on the second coin and the
tail on the first. Thus, the total number of equally likely ways is
4, and the probability of both coins showing heads is J^. The
probability of one head and one tail showing is %, so that a head
and a tail are twice as likely to appear as either two heads or 'two
tails.
Another example may prove useful. Suppose that a pair of
dice is thrown. What is the probability that a total of eight
shows? The total number of ways in which two dice can fall
is 36. "(This follows from the fundamental principle of com-
494 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §147
binatory analysis: if one thing can be done in m different ways and
another thing can be done in n different ways, then both things can
be done together, or in succession, in mn different ways.) The sum
of 8 can be obtained as follows: 2 and 6, 3 and 5, 4 and 4.
Now, there are two ways in which 2 and 6 can fall: 2 on the
first die and 6 on the second, and vice versa. Similarly, there are
two ways in which 3 and 5 can fall, but there is only one way in
which 4 and 4 can fall. Hence, the total number of equally likely
and favorable cases is 5, so that the desired probability is %e-
The two foregoing examples were solved simply by enumer-
ating all the possible and all the favorable cases. Frequently,
such enumeration is laborious and it is convenient to resprt to
formulas. Thus, let it be required to find the probability of
drawing 4 white balls from an urn containing 10 white, 4 black,
and 3 red balls. The number of ways in which 4 white balls
can be c'hosen from 10 white balls is equal to the number of com-
binations of 10 things taken 4 at a time, namely,
10!
lot-4 =
4!6!
The total number of ways in which 4 balls can be chosen from the
17 available is
_
17/4 ~ 4T13!'
Hence, the probability of success is
_ 10^4 _ 10! 4! 13! _ _8_
P ~~ 17C4~ 4! 6! 17! ~ 34'
Another example will illustrate further the use of formulas.
Suppose that it is desired to find the probability of drawing 4
white, 3 black, and 2 red balls from the urn in the preceding
illustration. In this case the number of ways in which 4 white
balls can be drawn is ioC4; the 3 black balls can be chosen in
4Ca ways; and the 2 red ones in 3^2 ways. An application of the
fundamental principle of combinatory analysis gives the required
probability as
ioC4 • 4C3 • 3C2 _ 252
P 17C9 ~ 2431'
§148 PROBABILITY 495
PROBLEMS
1. What is the probability that the sum of 7 appears in a single
throw with two dice? What is the probability of the sum of 1 1 ? Show
that 7 is the more probable throw.
2. An urn contains 20 balls: 10 white, 7 black, and 3 red. What is
the probability that a ball drawn at random is red? White? Black?
If 2 balls are drawn, what is the probability that both are white? If
10 balls are drawn, what is the probability that 5 are white, 2 black, and
3 red?
148. Independent Events. A set of events is said to be
independent if the occurrence of any one of them is not influenced
by the occurrence of the others. On the other hand, if the
occurrence of any one of the events affects the occurrence of the
others, the events are said to be dependent.
THEOREM 1. // the probabilities of occurrence of a set of n
independent events are pi, p2, • • • , pn, then the probability that
all of the set of events will occur is p = pipz • • * pn.
The proof of this theorem follows directly f eom the fundamental
principle. Thus, let there be only two events, whose probabilities
of success are
mi , w2
PI = TT and pz = -rp
The total number of ways in which both the events may succeed
is mim2, and the total number of ways in which these events can
succeed and fail to succeed is NiN*. Hence, the probability of
the occurrence of both of the events is
mi m2
p = NWl = Fi ' F2
Obviously, this proof can be extended to any number of events.
Illustration. A coin and a die are tossed. What is the
probability that the ace and the head will appear? The proba-
bility that the ace will appear is J^, and the probability that the
head will appear is J^. Therefore, the probability that both
head and ace will appear is
THEOREM 2. // the probability of occurrence of an event is p\
and if, after that event has occurred, the probability of occurrence
496 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §148
of a second event is pz, then the probability of occurrence of both
events in succession is p\p%.
The proof of this theorem is similar to that of Theorem 1 and
will be left to the student. Obviously, the theorem can be
extended to more than two events.
Illustration 1. What is the probability that 2 aces be drawn
in succession from a deck of 52 cards? The probability that an
ace will be drawn on the first trial is ^2- After the first ace
has been drawn, the probability of drawing another ace from the
remaining 51 cards is %i, so that the probability of drawing 2
aces is
Hi =
Illustration 2. What is the probability that the ace appears
at least once in n throws of a die? The probability of the ace
appearing in a single throw of the die is Y§, and the probability
that it will not appear is %. The probability that the ace will
not appear in n successive throws is
Hence, the probability that the ace will appear at least once is
Illustration 3. An urn contains 30 black balls and 20 white
balls. What is the probability that (a) A white ball and a
black ball are drawn in succession? (6) A black ball and a white
ball are drawn in succession? (c) Three black balls are drawn
in succession?
a. The probability of drawing a white ball is 2%$. After a
white ball is drawn, the probability of drawing a black ball is
3%g. Hence, the probability of drawing a white ball and a black
ball in the order stated is
P =
b. The probability of drawing a black ball is 3%o> and the
probability that the second drawing yields a white ball is
so that
P = 3%0 ' 2%9 = %•
c. The probability of drawing 3 black balls in succession is
P =
§149 PROBABILITY 497
Illustration 4. The probability that Paul will solve a problem
is Y±, and the probability that John will solve it is %. What is
the probability that the problem will be solved if Paul and John
work independently?
The problem will be solved unless both Paul and John fail.
The probability of John's failure to solve it is ^ and of PauPs
failure to solve it is %. Therefore, the probability that Paul
and John both fail is
M-H = H,
and the probability that the problem will be solved is
i - H = M-
PROBLEMS
1. What is the probability that 5 cards dealt from a pack of 52 cards
are all of the same suit?
2. Five coins are tossed simultaneously. What is the probability
that at least one of them shows a head? All show heads?
3. What is the probability that a monkey seated before a typewriter
having 42 keys with 26 letters will type the word sir?
4. If Paul hits a target 80 times out of 100 on the average and John
hits it 90 times out of 100, what is the probability that at least one of
them hits the target if they shoot simultaneously?
6. The probability that Paul will be alive 10 years hence is %, and
that John will be alive is %. What is the probability that both Paul
and John will be dead 10 years hence? Paul alive and John dead?
John alive and Paul dead?
149. Mutually Exclusive Events. Events are said to be
mutually exclusive if the occurrence of one of them prevents the
occurrence of the others. An important theorem concerning such
events is' the following:
THEOREM. The probability of the occurrence of either one or
the other of two mutually exclusive events is equal to the sum of the
probabilities of the single events.
The proof of this theorem follows from the definition of proba-
bility. Consider two mutually exclusive events A and B.
Inasmuch as the events are mutually exclusive, A and B cannot
occur simultaneously and the possible cases are the following: (a)
A occurs and B fails to occur, (6) B occurs and A fails to occur,
(c) both A and B fail. Let the number of equally likely cases in
which (a) A can occur and B fail be a, (6) B can occur and A fail
be 0, (c) both A and B fail be y.
498 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
Then the total number of equally likely cases is a + ft + y.
The probability that either A or B occurs is
a + /3
a + ft + y
the probability of occurrence of A is
a
a + ft + y
and the probability of occurrence of B is
Therefore, the probability of occurrence of either A or B is equal
to the sum of the probabilities of occurrence of A alone and of
B alone. Obviously, this theorem can be extended to any
number of mutually exclusive events.
The task of determining when a given set of events is mutually
exclusive is frequently difficult, and care must be exercised that
this theorem is applied only to mutually exclusive events. Thus,
in Illustration 4, Sec. 148, the probability that either Paul or
John will solve the problem cannot be obtained by adding
J^ and %, for solution of the problem by Paul does not elimi-
nate the possibility of its solution by John. The events in this
case are not mutually exclusive and the theorem of this section
does not apply.
Illustration 1. A bag contains 10 white balls and 15 black
balls. Two balls are drawn in succession. What is the proba-
bility that one of them is black and the other is white?
The mutually exclusive events in this problem are : (a) drawing
a white ball on the first trial and a black on the second; (6)
drawing a black ball on the first trial and a white on the second.
The probability of (a) is l%5 ' ^4 and that of (b) is ^5 • i%4,
so that the probability of either (a) or (6) is
Illustration 2. Paul and John are to ertgage in a game in
which each is to draw in turn one coin at a time from a purse
containing 3 silver and 2 gold coins. The coins are not replaced
after being drawn. If Paul is to draw first, find the probability
for each player that he is the first to draw a gold coin.
§149 PROBABILITY 499
The probability that Paul succeeds in drawing a gold coin
on the first trial is %. The probability that Paul fails and John
succeeds in his first trial is
The probability that Paul fails, John fails, and then Paul suc-
ceeds is
The probability that Paul fails, John fails, Paul fails again, and
John succeeds is
% • M • H • % = Ho,
for after three successive failures to draw a gold coin there
remain only the two gold coins in the purse and John is certain
to draw one of them. Therefore, Paul's total probability is
% + H = H
and John's probability is
Ho + Ho = %.
PROBLEMS
1. One purse contains 3 silver and 7 gold coins; another purse con-
tains 4 silver and 8 gold coins. A purse is chosen at random, and a
coin is drawn from it. What is the probability that it is a gold coin?
2. Paul and John are throwing alternately a pair of dice. The first
man to throw a doublet is to win. If Paul throws first, what is his
chance of winning on his first throw? What is the probability that
Paul fails and John wins on his first throw?
SfOn the average a certain student is able to solve 60 per cent of
the problems assigned to him, If an examination contains 8 problems
and a minimum of 5 problems is required for passing, what is the
student's chance of passing?
4. Two dice are thrown; what is the probability that the sum is either
Tor 11?
5. How many times must a die be thrown in order that the prob-
ability that the ace appear at least once shall be greater than % ?
6. Twenty tickets are numbered from 1 to 20, and one of them is
drawn at random. What is the probability that the number is a
multiple of 5 or 7? A multiple of 3 or 5?
Note that in solving the second part of this problem, it is incorrect to
reason as follows: The number of tickets bearing numerals that are
multiples of 3 is 6, and the number of multiples of 5 is 4. Hence, the
500 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §160
probability that the number drawn is either a multiple of 3 or of 5 is
+ ^o = «. Why?
publishing concern submits a copy of a proposed book to three
independent critics. The odds that a book will be reviewed favorably
by these critics are 5 to 4, 4 to 3, and 2 to 3. What is the probability
that a majority of the three reviews will be favorable?
8. If on the average in a shipment of 10 cases of certain goods 1 case
is damaged, what is the probability that out of 5 cases expected at least
4 will not be damaged?
150. Expectation. The expectation of winning any prize is
'defined as the value of the prize multiplied by the probability
of winning it. Let it be required to determine the price one
should pay for the privilege of participating in the following
game. A purse contains 5 silver dollars and 7 fifty-cent pieces,
and a player is to retain the two coins that he draws from the
purse. It can be argued fallaciously as follows: The mutually
exclusive cases are (a) 2 dollar coins, (6) 2 half-dollar coins, (c)
1 dollar coin and 1 half-dollar coin. Therefore, the values of
the prizes are $2, $1, and $1.50, and the fair price to pay for the
privilege of participating is $1.50. But, the probability of
obtaining (a) is
^ _ 5^2 _ 5
Pa ~ ^2 ~ 33'
that of obtaining (b) is
and that of obtaining (c) is
f^ 7
Pb = ^ = 22'
= 5.7 = 35
12(^2 66
Hence, the expectation of obtaining (a) is
6a = 2 • %3 = 0.30,
that of obtaining (b) is
e& = 1 • %2 = 0.32,
and that of obtaining (c) is
cc = 1.50 • 3%6 = 0.80.
It follows that the total expectation is
$0.30 + $0.32 + $0.80 = $1.42,
instead of $1.50.
§151 PROBABILITY 501
PROBLEMS
1. A batch of 1000 electric lamps is 5 per cent bad. If 5 lamps are
tested, what is the probability that no defective lamps appear? What
is the chance that a test batch of 5 lamps is 80 per cent defective? What
is a fair price to pay for a batch of 500 such lamps if the perfect ones can
be bought for 10 cts. each?
2. What is a fair price to pay for a lottery ticket if there are 100
tickets and 5 prizes of $100 each, 10 prizes of $50 each, and 20 prizes of
$5 each?
3. What is the expected number of throws of a coin necessary to
produce 3 heads?
151. Repeated and Independent Trials. Frequently it is
required to compute the probability of the occurrence of an
event in n trials when the probability of the occurrence of that
event in a single trial is known. For example, it may be required
to find the probability of throwing exactly one ace in 6 throws
of a single die. The possible mutually exclusive cases are as
follows:
(1) An ace on the first throw, and none on the remaining 5
throws.
(2) No ace on the first throw, an ace on the second, and no
aces on the remaining 4 throws.
(3) No ace on the first 2 throws, an ace on the third, and no
aces on the last 3 throws.
(4) No aces on the first 3 throws, an ace on the fourth, and no
aces on the last 2 throws.
(5) No aces on the first 4 throws, an ace on the fifth, and no
ace on the last throw.
(6) No aces on the first 5 throws, and an ace on the last throw.
The probability of the occurrence of (1) is
since the probability of throwing an ace on the first trial is %
and the probability that the ace will not appear on the succeeding
5 throws is (%)5. The probability of (2) is
the probability of (3) is
502 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §151
and it is clear that the probability of any one of the 6 specified
combinations is
Since the cases are mutually exclusive, the probability that some
one of the 6 combinations will occur is
P = 6 - H - W = (%)5«
It should be observed that the probability of obtaining any
combination of 1 ace and 5 not-aces is always the same, so that in
order to obtain the probability of occurrence of one of the set of
mutually exclusive cases all that is necessary is to multiply the
probability of the occurrence of any specified combination by the
number of distinct ways in which the events may occur. This
leads to the formulation of an important theorem which is
frequently termed the binomial law.
THEOREM 1 (Binomial Law). // the probability of occurrence
of an event in a single trial is p, then the probability that it will occur
exactly r times in n independent trials is
Pr = nCrpr(l ~ pY" '
where
c - -iL
nCr ~ r\(n - r)!'
The method of proof of this theorem is obvious from the
discussion of the specific case that precedes the theorem. The
probability that an event will occur in a particular set of r
trials and fail in the remaining n — r trials is pr(l — p)n~r.
But since the number of trials is n, the number of ways in which
the event can succeed r times and fail n — r times is equal to the
number of the combinations of n things taken r at a time, or
n\
nv-T —
r\(n — r)!
Hence the probability of exactly r successes and n — r failures is
(151-1} P' - r\(n-r)\ ^ ~ ">~
Illustration 1. What is the chance that the ace will appear
exactly 4 times in the course of 10 throws of a die?
§161 PROBABILITY 503
Formula (151-1) gives
_ 10! /1V M" _ 656,250
_
4 ~ 4l6! \fi ~ 60,466,170 ~
Illustration 2. Ten coins are tossed simultaneously. What is
the chance that exactly 2 of them show heads?
Here
If in the example at the beginning of this section it had been
required to determine the probability that the ace would appear
at least once in the course of the 6 trials, the problem would be
solved with the aid of the following argument: The ace will
appear at least once if it appears exactly once, or exactly twice,
or exactly three times, and so on. But the probability that it
appears exactly once is
PI = .C
exactly twice,
p2 = 6
exactly three times,
etc. These compound events are mutually exclusive, so that the
probability of the ace appearing at least once is the sum of the
probabilities
Pi, P2, Pa, ' ' ' , Pe.
The general theorem, which includes this problem as a special
case, is the following.
THEOREM 2. // the probability of the occurrence of an event
on a single trial is p, then the probability that the event will occur
at least r times in the course of n independent trials is
P>r = Pn + nClpn~lq + *C2p»-2 g2 + • • • + nCn-rprqn-r,
where q = 1 — p.
It should be noted that nCr = nCn-r is the coefficient of pr in the
binomial expansion for (p + g)n and that p^r is equal to the sum
of the first n — r + 1 terms in the expansion for (p + q)n.
Illustration 3. The probability that at least 2 of the coins
show heads when 5 coins are tossed simultaneously is
= PAY + 6
504 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §162
The first term of this sum represents the probability of exactly 5
heads, the second represents that of exactly 4 heads, the third
that of exactly 3 heads, and the last represents that of exactly
2 heads.
PROBLEMS
1. If 5 dice are tossed simultaneously, what is the probability that
(a) exactly 3 of them turn the ace up? (6) At least 3 turn the ace up?
2. If the probability that a man aged 60 will live to be 70 is 0.65, what
is the probability that out of 10 men now 60, at least 7 will live to be 70?
3. A man is promised $1 for each ace in excess of 1 that appears in 6
consecutive throws of a die. What is the value of his expectation?
4. A bag contains 20 black balls and 15 white balls. What is the
chance that at least 4 in a sample of 5 balls are black?
5. Solve Prob. 3, Sec. 149.
162. Distribution Curve. Some interesting and useful con-
clusions can be deduced regarding the formula for repeated
independent events from the consideration of an example that
presents some features of the general case. Consider a purse
in which are placed 2 silver and 3 gold coins, and let it be required
to determine the probability of drawing exactly r silver coins
in n repeated trials, the coin being replaced after each drawing.
The probability of exactly r successes in n trials is given by the
binomial law [see (151-1)]
Pr = nCrpr(l ~ p)***,
where p, the probability of drawing a silver coin on a single trial ,
is %.
If the number of drawings is taken as n = 5, the probability
that none of the drawings yields a silver coin is
Po = 6Co(%)0(9s)6 = 0.07776,
the probability that 5 trials yield exactly 1 silver coin is
Pi = 6<7i(%)(^)4 = 0.2592,
and the probability that exactly 2 silver coins will appear is
P2 = 6C2(%)2(%)3 = 0.3456.
In this manner, it is possible to construct a table of the values
that represent the probabilities of drawing exactly 0, 1, 2, 3, 4, 5
silver coins in 5 trials. Such a table, where the values of pr are
computed to four decimal places, is given next.
PROBABILITY 505
PROBABILITY OF EXACTLY r SUCCESSES IN 5 TRIALS
r
Pr
r
Pr
0
0 0778
3
0 2304
1
0 2592
4
0 0768
2
0 3456
5
0 0102
It will be observed that r = 2 gives the greatest, or "most
probable," value for pr, which seems reasonable in view of the
fact that the probability of drawing a silver coin on a single trial
is % and one would " expect" that 2 silver coins should result
from 5 repeated drawings.
If the number of trials is n = 10, the formula
Pr = lo
gives the following set of probabilities for 0, 1, 2, 3, • • • , 10
successes.
PROBABILITY OF EXACTLY r SUCCESSES IN 10 TRIALS
r
Pr
r
Pr
r
Pr
0
0 0060
4
0 2508
8
0 0106
1
0 0403
5
0 2007
9
0 0016
2
0 1209
6
0 1115
10
0 0001
3
0 2150
7
0 0425
Again it appears that the most probable number of successes in n
trials is equal to the probability of success in a single trial
multiplied by the number of trials.
If a similar table is constructed for n = 30, the resulting proba-
bilities are as shown below.
PROBABILITY OF EXACTLY r SUCCESSES IN 30 TRIALS
r
Pr
r
Pr
r
Pr
<2
0 0000
9
0 0823
16
0 0489
3
0 0003
10
0 1152
17
0 0269
4
0 0012
11
0 1396
18
0 0129
5
0 0041
12
0.1474
19
0 0054
6
0 0115
13
0 1360
20
0 0020
7
0 0263
14
0 1100
21
0 0006
8
0.0505
15
0 0783
22
0 0002
^23
0 0000
506 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §152
In this table the entry 0.0000 is made for 0 < r < 2 and for
r > 23 because the values of pr were computed to four decimal
places, and in these cases pr was found to be less than 0.00005.
For example, the probability of drawing exactly 1 silver coin in
30 trials is
Pi = 30(%)(%)29 = 0.00000442149,
and the probability of drawing exactly 23 silver coins in 30 trials
is
p23 = 3oC23(%)23(%)7 = 0.000040128.
Therefore, for all values 23 < r < 30, the values of pr are less
than 0.00005 and must be recorded as 0.0000.
Just as in the foregoing tables, the most probable number is
equal to %n> although the probability of drawing exactly 12
Pr
04
03
~-_
n-5 0.4
03
n-10
02
02
0 1
(
02
i.
-.1 i.
)
2 3 4 5 r 0123456789 10 r
01
: ..rill 1
0 2 4 6 8 10 12 14 16 18 20 22 24
FIG. 148.
silver coins, 0.1474, is less than the probability of drawing the
most probable number of silver coins in the case of 10 trials.
These tabulated results can be more conveniently represented
in a graphical form, where the values of r are plotted as abscissas
and the values of pr as ordinates. Such graphs are known as
distribution charts (Fig. 148).
An alternative method of graphical representation is obtained
by erecting rectangles of unit width on the ordinates which
represent the probabilities pr of occurrence of r successes. Since
the width of each rectangle is unity, its area is equal, numerically,
to the probability of the value of r over which it is erected. In
such graphs the vertical lines are not essential to the interpreta-
tion of the graph and hence are omitted. The resulting broken
curve constitutes what is known as a distribution curve. The
§152
PROBABILITY
507
area under each step of the curve represents the value of pr;
and the entire area under the distribution curve is unity, for it
represents the sum of the probabilities of 0, 1, 2, • • • , n suc-
cesses. Such curves, corresponding to the distribution charts
(Fig. 148), are drawn in Fig. 149.
It appears that, as the number of drawings, n, is increased, the
probability of obtaining the most probable number of silver coins
decreases. Moreover, there is a greater spread of the chart as
the number of trials is increased, so that the probability of missing
the most probable number by more than a specified amount
increases with the increase in the number of drawings.
0.4
n»5
0.3
-
02
0.1
f
I.
03
02
0 1
n-10
012345
01 23456789 10
n-30
10 12 14 16
FIG. 149.
18 20 2Z 24 26 r
The following observations will serve to clarify the last state-
ment. In the case of 5 trials the probability of missing the
most probable number of successes by 5 is zero, for the deviation
from the most probable number 2 cannot be greater than 3.
But in the case of 10 trials the probability of missing the most
probable number by this same amount has a definite non-zero
value. Thus, in order to miss the most probable number 4
by 5, r must be either 9 or 10. Hence, the probability of missing
4 is 0.0016 + 0.0001 = 0.0017. In the case of 30 trials the
most probable number of successes is 12; and, in order to miss
12 by 5, r must be less than 8 or greater than 16. Therefore the
probability of missing 12 by 5 is the sum
30
X Pr + X Pr = 0.0914.
If the number of drawings n is made 1,000,000, the most probable
508 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §163
number of successes is 400,000, and the probability of missing it
by 5 is very nearly unity. On the other hand, the probability
of obtaining the most probable number 400,000 is a very small
quantity.
The important facts obtained from the foregoing considerations
are the following:
1. The most probable number of successes appears to be equal
to pn.
2. The probability of obtaining the most probable number of
successes decreases with the increase in the number of trials n.
3. The probability of missing the most probable number by a
specified amount increases with the increase in the number of
trials.
It can be established that the last two facts, inferred from the
special example, are true in general. The first fact, concerning
the size of the most probable number, clearly is meaningless if
pn is not an integer. Thus, if the number of drawings is n — 24
and p = %, then pn — 4%. It can be shown in general that the
most probable number is pn, provided pn is an integer; otherwise,
the most probable number is one of the two integers between
which pn lies. In fact, the following is the complete statement
of the theorem:* The most probable number of successes is the
greatest integer less than np -{-p. If np + p is an integer, there
are two most probable numbers, namely, np + P and np + p — 1.
Since p < I, it is clear that the most probable number of successes
is approximately equal to np. This number np is called the
"expected" number of successes.
PROBLEM
A penny is tossed 100 times. What is the most probable number of
heads? What is the probability of this most probable number of
heads? If the penny is tossed 1000 times, what is the probability of the
most probable number of heads?
153. Stirling's Formula. The binomial law (151-1), on which
the major portion of the theory of probability is based, is exact,
but it possesses the distinct disadvantage of being too com-
plicated for purposes of computation. The labor of computing
the values of the factorials that enter in the term nCr becomes
* For proof and further discussion see T. C. Fry, Probability and Its Engi-
neering Uses, Chap. IV.
§153
PROBABILITY
509
prohibitive when n is a large number. Accordingly, it is desirable
to develop an approximation formula for n!, when n is large.
An asymptotic formula, which furnishes a good approximation
to n!, was developed by J. Stirl- y t
ing. By an asymptotic formula
is meant an expression such
that the percentage of error
made by using the formula as
an approximation to n \ is small
when n is sufficiently large, j k-1 k
whereas the error itself increases
with the increase in n. It will be Fl0' 15°*
indicated that for values of n greater than 10 the error made in
using Stirling's formula*
(153-1) n\ ~ nne~n
is less than 1 per cent.
Consider the function y = log x, and observe that, for k
log x dx > HHog (k - 1) + log fc],
2,
since the right-hand member represents the trapezoidal area
formed by the chord (Fig. 150) joining the points P and Q on the
curve y — log x. Denote the area between the chord and the
curve by a*, so that
+ an).
(153-2) J;fc_i log x dx = ^[log (fc - 1) + log k] + ak.
Setting k = 2, 3, • • • , n in (153-2) and adding give
| log x dx = K(log 1 + log 2) + H(l°g 2 + log 3) +
+ /^[log (n — 1) + log n] + (a2 + a3 + • • •
Integrating the left-hand member and combining the terms of the
right-hand member give
n
n log n — n + 1 = log n\ — Yi log n + ^ o».
Hence,
(153-3) log n! = (n + ^) log n - n + 1 - ^ at.
* The symbol ~, which is read " asymptotically equal to," is used instead
of = to call attention to the fact that the formula is asymptotic.
510 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §153
Since each at is positive, it follows that
log nl < (n + Y^) log n - n + 1,
and hence
(153-4) nl < e \/n nne~n.
The expression on the right of the inequality (153-4) is, therefore,
an upper bound for nl
To get a lower bound, solve (153-2) for «&, perform the integra-
tion, and obtain
(153-5) a*
Now, since the integrand is non-negative,
f* A iY
(153-6) (£ - f ) dx > 0,
Jk-i \x k)
and the evaluation of (153-6) leads to the formula
10 _*_ < 2k ~ l
Ki J. juK/\fc L)
By the use of this inequality, (153-5) gives
1 1 /_!
ak 4k(k ~- 1) 4 \A; - 1
Hence,
By means of this result and (153-3), one obtains
log nl > (n + Yd log n - n + 1 - y±,
whence
(153-7) nl > e^ \/n nne~n.
Combining (153-4) and (153-7) furnishes the inequality*
e^ \/n nne~n < nl < e -\/n nne~n,
for all values of n > 1. Since e = 2.718, e* = 2.117, and
\/25r = 2.507, it follows that
n I c^ nne~n \/27rn.
* The derivation of this result is given by P. M. Hummel, in Amer. Math.
Monthly, vol. 47, p. 97, 1940.
§154 PROBABILITY 511
It is possible to obtain a sharper lower bound for n\ by. using
the integral*
(153-8)
'
f* [-~
j^-i L£
instead of (153-6).
To gain some insight into the accuracy of this formula, note
that (153-1) gives for n = 10 the value 3,598,696, whereas the
true value of 10! is 3,628,800. The percentage of error in this
case is 0.8 per cent. Forn = 100, (153-1) gives 9.524847 X 10157,
whereas the true value of 100! is 9.3326215 X 10157, so that the
percentage of error is 0.08. It is worth noting that, even for
n = 1, the error is under 10 per cent and that for n = 5 it is in
the neighborhood of 2 per cent.
PROBLEM
Make use of (153-8) in order to show that
e'Ma -y/^ nne~n < n!, if n > 1,
and compare the value of e1^2 with that of \/2ir.
154. Probability of the Most Probable Number. It was
mentioned in Sec. 152 that the most probable number of successes
is either equal, or very nearly equal, to the expected number
e = np. Very often it is desirable to compute the probability
of the expected or the most probable number of successes. Of
course, p€ can be computed from the exact law by substituting
in it r = np, but formula (151-1) is cumbersome to use when
factorials of large numbers appear in nCr. An approximate
formula can be obtained by replacing n\ and (np)! by their
approximate values with the aid of Stirling's formula. It is
readily verified that, when these replacements are made, the prob-
ability of the most probable number of successes is approximately
where q — 1 — p. It must be kept in mind that (154-1) is
subject to the same restrictions as (153-1) and gives good results
for np ^ 10.
* See problem at the end of this section.
512 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §155
Thus, if a die is tossed 100 times, the most probable number of
aces is 16. The exact formula gives
whereas the approximation (154-1) gives
p" = = °-309-
The percentage of error is quite small.
PROBLEMS
1. Two hundred and fifty votes were cast for two equally likely can-
didates for an office. What is the probability of a tie?
2. What is the most probable number of aces in 1200 throws of a die?
What is the probability of the most probable number?
3. Solve, with the aid of the approximate formula, the problem at the
end of Sec. 152.
155. Approximations to Binomial Law. With the aid of
formula (153-1), it is possible to devise various formulas approxi-
mating the binomial law (151-1). One of these approximations
is known as the Poisson formula or the law of small numbers.
The wide range of applicability of this law can be inferred from
the fact that it has been used successfully in dealing with such
problems as those of beta-ray emission, telephone traffic, trans-
mission-line surges, and the expected sales of commodities. The
law of small numbers gives a good approximation to (151-1) in
those problems in which r is small compared with the large number
n, and p represents the probability of occurrence of a rare event in a
single trial.
Replacing n\ and (n — r)! in (151-1) with the aid of Stirling's
formula (153-1) leads to
uss-n 0 ~ o^i - «>
(1551) p,____p(l p)
= —, V-r+H/' ^
4 ~ ;)
By hypothesis, r is small compared with n, so that
§155 PROBABILITY 513
is very nearly equal to
which, * for large values of n, differs little from e~r. Similarly,
(1 _ p)»-r - (1 _ p)^
which in turn is nearly equal to e~Mp, since
(1 - p)» = 1 - np + ^~^-} p» - • • •
and
«o 2/rj2
e-«P = 1 - np + -~2y - • • • .
The substitution of e~r for
and e~np for
in (155-1) leads to the desired law of small numbers,
(\ tt 9"\ T> — ^nV) p—np
(LOO-*) Pr — rj e ".
Formula (155-2) is frequently written in a slightly different
form. It will be recalled that the expected number of successes
is e = np, so that (155-2) can be written
= 6^ _e
r\
An application of this law to some specific cases may prove
interesting. Suppose that it is known that, on the average, in a
large city two persons die daily of tuberculosis. What is the
probability that r persons will die on any day? In this case the
expected number of deaths is c = 2, so that
r = - e~2
r\
* Note that lim (1 + l/n)rt = e. For a rigorous discussion, see I. S.
n— * oo
Sokolnikoff, Advanced Calculus, pp. 28-31.
514 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §165
Therefore,
r
Pr
r
Pr
r
Pr
0
0 136
2
0 272
4
0 091
1
0 272
3
0 181
5
0 036
A glimpse into the accuracy of this law can be gained by con-
sidering the following example.
Example. What is the probability that the ace of spades will be
drawn from a deck of cards at least once in 104 consecutive trials? This
problem can be solved with the aid of the exact law (151-1) as follows:
The probability that the ace will not be drawn in the 104 trials is
Po =
= 0.133,
and the probability that the ace will be drawn at least once is 1 — 0.133
= 0.867. On the other hand, Poisson's law (155-2) gives for the prob-
ability of failure to draw the ace
e ' — f/ .
/'O —
Hence, the probability of drawing at least one ace of spades is 1 — e~*
= 0.865.
Another important approximation to the binomial law (151-1),
namely,
n\
(155-3)
Pr =
r\(n - r)!
prqn
where q = 1 — p, is obtained by assuming that r, n, and n — r
are all large enough to permit the use of the Stirling formula.
Replacing n!, r!, and (n — r)! by Stirling's approximations
gives, upon simplification,
r(n - r)
Let 5 denote the deviation of r from the expected value np;
that is,
8 = r — up.
Then,
n — r = nq — 6,
§155 PROBABILITY 515
and (155-4) becomes
\ f
or
v— (np+8)
(, \
i + JL)
np/
where
A =
Then,
log prA = -(np + 6) log ( 1 + — J - (nq - 5) log ( 1 - — -V
Assuming that \d\ < npq, so that
< 1 and
np
nq
permits one to write the two convergent series
log (i + A) = A - * + *'
\ nP/ nP 2n2p2 3r
and
Hence,
logy
2npq
Now, if \d\ is so small in comparison with npq that one can
neglect all terms in this expansion beyond the first and can
replace A by \/2irnpq, then there results the approximate
formula
(155-5) Pr = * e~^*,
\f2irnpq
which bears the name of Laplace's, or the normal, approximation.
Since the maximum value of the exponential e~x, for x ^ 0, is
unity, it follows that the normal approximation gives for the
probability that r will assume its most probable value the same
516 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §156
value as was obtained in Sec. 154. It is obvious that the normal
approximation gives best results when p and q are nearly equal.
If the mean error a is defined
by the formula
0- = \/npq,
then (155-5) assumes the form
151.
pr =
I
- e
and the graph of pr as a function of d is a bell-shaped curve
(Fig. 151), known as the normal distribution curve.*
PROBLEMS
1. What is the probability of throwing an ace with a die exactly 10
times in 1200 trials?
2. A wholesale electrical dealer noticed that a shipment of 10,000
electric lamps contained, on the average, 20 defective lamps. What is
the probability that a shipment of 10,000 lamps is 1 per cent defective?
3. In a certain large city, on the average, two persons die daily of
cancer. What is the probability of no persons dying on any day? One
person dying? Two? Three? Four? Five?
4. Two dice are tossed 1000 times. What is, approximately, the
probability of getting a sum of 4 the most probable number of times?
5. What is the approximate probability that a sum of 4 will appear
500 times in a set of 1000 tosses?
156. The Error Function. Let wi, ra2, • • • , mn be a set of n
measurements, of some physical quantity, that are made inde-
pendently and that are equally trustworthy. If the best estimate
of the value of the measurements is m, then the " errors" in
individual measurements are
— m,
xn = mn — m,
Xi = mi — m, #2 =
and the sum of the errors is
(156-1) xi + Xz + • • • + xn = (mi + m2 + • • • + mn) - mn.
If it is assumed that on the average the positive and negative
errors are equally balanced, then their sum is zero, and (156-1)
becomes
mn = mi + w2 + • • • + mn
* For a detailed discussion see T. C. Fry, Probability and Its Engineering
Uses.
t=» 1
§158 PROBABILITY 517
or
(156-2) m =
IV
It is important to note that the best value m is the arithmetic
average of the individual measurements when it is assumed that
the positive and negative errors are equally likely. In per-
forming a set of measurements, not all the errors xl are equally
likely to occur. In general, large errors are less likely to occur
than small ones. For instance, the probability of making an
error of 1 ft. in measuring the length of a table is less than that
of making an error of 1 in.
Let the probability of making an error xl be denoted by
<p(xi). The assumption that positive and negative errors are
equally likely to occur de-
mands that * x
which states that <p(x) is an
even function. Further-
more, the hypothesis that
small errors are more likely
to occur than large ones re-
quires <p(x) to be a decreasing function for x > 0; and since
infinitely large errors cannot occur,
*(«>) = 0.
These observations lead to the conclusion that the function
<p(x), which gives the probability of occurrence of the error #,
must have the appearance shown in Fig. 152, where the errors xl
are arranged in order of increasing magnitude. Upon recalling
the fact that the ordinates represent the probability of occurrence
of an error of any size x, it is clear that the area under this curve
from — oo to + oo must be unity, for all the errors are certain
to lie in the interval (— <», oo). Hence,
/as
<p(x) dx = 1.
- °°
Moreover, the probability that the error lies between the limits
Xt and £t + A# is equal to the area bounded by curve y = <p(x),
518 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §166
the ordinates y = xl and y = xl + A#, and the x-axis, which is
equal to the value of the integral
r
dx.
By hypothesis the measurements rat were made independently,
so that the probability of simultaneous occurrence of the errors
Xi, xt, • • • , xn is equal to the product of the probabilities of
occurrence of the individual errors, or
(156-3) P = <P(XI}<P(X*} • • • <p(xn)
= <p(m\ — m)(p(mz — m) • • • <p(mn — m).
The expression (156-3) is a function of the best value m, in which
the functional form of <p is not known. Now, if it be assumed
that the best value m is also the most probable value, that is, the
value which makes P a maximum, then it is possible to determine
the functional form of <p by a method due to Gauss. In other
words, it is taken as a fundamental axiom that the probability
(156-3) is a maximum when m is the arithmetic average of the
measurements mi, m^ • • • , mn> But if (156-3) is a maximum,
its logarithm is also a maximum. Differentiating the logarithm
of (156-3) with respect to m and setting the derivative equal to
zero give
M r,A A\ <i - m) <p'(mz - m) <p'(mn ~ m)
( 1 00-4 ) — , --- r H -- -f -- r- + ' ' ' H -- y -- r = 0.
<f>(m>i ~ in) <p(mz — m) <p(mn — m)
n
This equation is subject to the condition 2 a:t = 0.
t = i
If
<p'(mt — m)
<p(mt — m)
is set equal to F(xt), (i = 1, 2, • • • , n), Eq. (156-4) can be
written as
(156-5) F(XI) + F(x*) + • - - + F(xn) = 0
n
with S #t = 0. If there are only two measurements, (156-5)
reduces to
= 0,
§166 PROBABILITY 519
with Xi + xz = 0, or #2 = — #i. Therefore,
F(xi)+F(-xi) = 0,
or
(156-6) F(x) = -F(-x).
Similarly, if there are only three measurements, then
Ffa) + F(xt) + F(x9) = 0,
with xi + £2 + #3 = 0. Therefore,
But, from (156-6),
and since — x$ — x\ + #2,
F(*0 + f (s,) = F(Xl + x,).
Differentiating this expression partially with respect to x\ and
leads to
F'(xJ = F^rci + x>) and
or
(156-7) F'(xi) = F'(x*).
Since x\ and x% are independent, (156-7) can be true only if
*"(a*) = c,
so that
F(XI) = cxi and F(x*) = co;2.
Recall that, by definition,
so that the differential equation for <p is
which, upon integration, gives
(156-8) q>(x)
where K and c are arbitrary constants.
520 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §156
One of these constants can be determined at once, for it is
known that
*(x) dx = 1.
The substitution of <p(x) in this integral gives
K f °° e~h^dx = 1,
J— oo
where —h2^ c/2.
This integral can be evaluated by means of a procedure similar
to that used in Sec. 81. Set
and then
f°° >j r *s
I* = e~* dx e-y dy
Jo Jo
= f f e-^^dxdy
Jo Jo
rl r
I I
Jo Jo
w
e~r~r dr d<p = 7;
4
where the last step results from the transformation of the double
integral into polar coordinates and has been described in Sec.
81. Hence,
f* °° /
I = e~*z dx = ¥!•
Jo *
But
K f* e-^dx = 2K f°° e~h^ dx = 1,
J- oo JO »
so that
1 J, Z,
X =,-
e~h2** dx 2 e~h2*2 d(hx)
Thus, (156-8) can be written as
(156-9) v(x] -**
which is called the Gaussian law of error. The undetermined
constant A, as will be seen in the next section, measures the
accuracy of the observer and is known as the precision constant.
§157
PROBABILITY
521
It is easy to verify the fact that the choice of <p, specified by
(156-9), gives a maximum for the product (156-3) when the sum
of the squares of the errors is a minimum. In fact, since #» =
ml — my (156-3) becomes
._ f * Y
-h* 2-
and the maximum value of P is clearly that which makes the
sum of the squares of the errors a minimum.
In order to verify the assumption that the choice of the
arithmetic average for the best value leads to the least value for
the sum of the squares of the
errors, all that is necessary is
to minimize
x,2 - 2) (ro.
The theory of errors based
upon the Gaussian law (156-9)
is often called the theory of
least squares,
157. Precision Constant.
Probable Error. In the pre-
ceding section, it was established that the probability of commit-
ting an error of magnitude x is given by the ordinate of the curve
FIG. 153.
- e~
y =
This curve is called the probability curve. Clearly, the proba-
bility of an error lying in the interval between x — — € and
x — +e is equal, numerically, to the area bounded by the proba-
bility curve (Fig. 153), the ordinates x = — € and x = +«, and
the x-axis. If only the absolute value of the error is of interest,
then the probability that the absolute value of the error does not
exceed € is
2h /»•
V^ Jo
e~h*x* dx.
If hx is set equal to t, this integral assumes the form
522 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §157
2 /•*•
= ^Jo '
which shows that P is a function of Ae, and, for a fixed value of
€, P increases with A. For large values of A, the probability
curve decreases very rapidly from its maximum value, h/\/w
at x = 0, to very small values, so that the probability of making
large errors is very small. On the other hand, if h is small, the
probability curve falls off very slowly so that the observer is
almost as likely to make fairly large errors as he is to make
small ones. For this reason the constant A is known as the pre-
cision constant.
That particular error which is just as likely to be exceeded as
riot is called the probable error. More precisely, the probable
error is that error e which makes P = }^, or
e~tz dt.
An approximate solution of this equation can be obtained by
expanding e~tz in Maclaurin's series, integrating the result term
by term, and retaining only the first few terms of the resulting
series.* The solution, correct to four decimal places, found by
this method is
he = 0.4769,
so that the probable error is 0.4769/A. It is commonly denoted
by the letter r.
In addition to the probable error, the mean absolute error and
the mean square error are of importance in statistics. The mean
absolute error is defined as
XOO
xe-hW fa _
0.5643
I <vr> — /*-*- fiw — —
v£,
and the mean square error is defined as
_. __ 2h
x& -- —
V*
C °° 2 fc,x, 7 _ 1
• x e ax — ^Tg-
Jo 2h
It will be observed that the mean absolute error is the z-coordi-
nate of the center of gravity of the area bounded by the proba-
*See Prob. 3, at the end of this section.
§167 PROBABILITY 523
bility curve and the positive coordinate axes, and that the
square root of the mean square error is the radius of gyration of
that area about the y-axis.
The values of these mean errors can actually be computed for
any set of observations. Thus,
wt — m
11 n n
so that
(157-1) h = -
P
Also,
n
V (m, - m
n "n 2h2'
so that h computed from this equation is
(157-2) h = — J— r
Vz2 \/2
These two expressions for the precision constant give a means of
computing h for any set of observation data. The two values of
h cannot be expected to be identical; but unless there is a fair
agreement between them, experience indicates that the data are
not reliable. The value of -y/x2 = °" *s commonly called the
standard deviation, and it follows from the foregoing that the
probable error is equal to 0.6745<r.
PROBLEMS
1. Evaluate the integral J Q e~^ dt by expanding the integrand in
series, and show that
f*X r3 r5 r7 T9
jC^-'-^n + s^-T^ + Ai-*
where R < a?u/1320.
2. The expression for the probability integral given in the preceding
problem is not suitable for computation purposes when x is large. But
524 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §157
Cx e-t* dt = f * er<2 dt - f °° er<2 <&
JO JO Jx
Show, by integrating by parts, that
f* 2J fl SJ e-*Yi ]
Jx '-"*-J, F'-'^-W^-S-'
and thus obtain an asymptotic expansion for the probability integral
that can be used to compute its value when x is large. Also, show that
the asymptotic series
r* , i V* e-*2r 1,1-3 1-3-5, n
Jo ^^^-^L1"^ w~ wr + • ' * J
gives a value for the integral which coffers from its true value by less
than the last term which is used in the series.
3. Show, with the aid of Homer's method, that the value of the
probable error is
0.4769
4. Compute the probable errors for the following set of observation
data:
mi = 1.305, ra2 = 1.301, w3 = 1.295, m* = 1.286,
m5 = 1.318, w6 = 1.321, ra7 = 1.283, w8 = 1.289,
m9 = 1.300, WIQ = 1.286,
by using (157-1) and (157-2).
5. With reference to Prob. 4, what is the probability of committing an
error whose absolute value is less than 0.03?
6. Two observers bring the following two sets of data, which repre-
sent measurements of the same quantity:
(a) mi = 105.1,
mB = 104.8,
(6) mj = 105.3,
m5 = 106.7,
m2 = 103.4,
m6 = 105.0,
m2 = 105.1,
m6 = 102.9,
m3 = 104.2,
m7 = 104.9.
m3 = 104.8,
m7 = 103.1.
m4 - 104.7,
m4 = 105.2,
Which set of data is the more reliable?
7. Discuss the problem of a rational way of proportioning the salaries
of two observers whose precision constants are hi and hz.
CHAPTER XII
EMPIRICAL FORMULAS AND CURVE FITTING
An empirical formula is a formula that is inferred by some
scheme in an attempt to express the relation existing between
quantities whose corresponding values are obtained by experi-
ment. For example, it may be desired to obtain the relation
connecting the load applied to a bar and the resulting elongation
of Hie bar. Various loads are applied, and the consequent
elongations are measured. Then, by one of the methods to be
given in this chapter, a formula is obtained that represents the
relationship existing between these two quantities for the
observed values. With certain restrictions, this formula can
then be used to predict the elongation that will result when an
arbitrary load is applied.
It is possible to obtain several equations of different types
that will express the given data approximately or exactly.
The question arises as to which of these equations will give the
best "fit" and be most successful for use in predicting the results
of the experiment for additional values of the quantities involved.
If there are n sets of observed values then, theoretically at least,
it is possible to fit the given data with an equation that involves
n arbitrary constants. What would be the procedure if it were
desired to obtain an equation representing these data but
involving less than n arbitrary constants? Questions of this
type will be considered in the succeeding sections.
158. Graphical Method. The graphical method of obtaining
an empirical formula and curve to represent given data is prob-
ably already somewhat familiar to the student from elementary
courses. It is particularly applicable when the given data can
be represented by equations of the three types
(1) y = mx + 6; (2) y = a + bxn; (3) y = kamx.
If the corresponding values (x*, yv) of the given data are
plotted on rectangular coordinate paper and the points thus
525
526 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §168
plotted lie approximately on1 a straight line, it is assumed that
the equation
y = mx + b
will represent the relationship. In order to determine the values
of the constants m and 6, the slope and ^-intercept may be read
from the curve, or they can be determined by solving the two
simultaneous equations
y\ = mxi + b, 2/2 = mx* + b
obtained by assuming that any two suitably chosen points
(xi, ?/i) and (#2, 2/2) lie on the line. Obviously the values of m
and b will depend upon the judgment of the investigator regard-
less of which method is used for their determination.
Consider the equation
y = a + bxn.
If the substitution xn = t is made, then the graph of y = a + bt
is a straight line and the determination of a and b is precisely
the same as in the preceding case. In the special case y = bxn,
taking logarithms on both sides gives
log y = log b + n log x,
which is linear in log y and log x and gives a straight line on
logarithmic paper. The slope of this line and the intercept on the
log i/-axis can be read from the graph. Hence, if the correspond-
ing values (#t, ?/,), when plotted on logarithmic paper, give points
that lie approximately on a straight line, the data can be repre-
sented by the equation
y = bxn,
whose constants can be read from the graph.
Similarly, if the data can be represented by a relation
y = kamxj
the corresponding values, when plotted on semilogarithmic*
paper, will give points that lie approximately on a straight line.
For taking logarithms on both sides of this equation gives
log y = log k + (m log a)x,
* For a discussion of logarithmic and semilogarithmic paper, see C. S.
Slichter, Elementary Mathematical Analysis.
§159
EMPIRICAL FORMULAS AND CURVE FITTING
527
which is linear in log y and x and therefore plots as a straight line
on semilogarithmic paper.
The three types of equations cited here are, of course, not the
only ones to which the graphic method is applicable. However,
they are the simplest because of the fact that their graphs, on
appropriate paper, give straight lines. When the points repre-
senting the observed values do not approximate a straight line,
some other method is usually preferable.
PROBLEMS
1. Find the equation that represents the relation connecting x and
y if the given data are
X
•
3
4
5
6
7
8
9
10
11
12
y
5
5 6
6
6 4
7
7 5
8 2
8 6
9
9 5
2. Find the equation of the type y — bxn that represents the relation
between x and y.
X
i
2
3
A
5
6
7
8
9
y
2 5
3 5
4 3
5
5 6
6 2
6 6
7 1
7 5
3. From the following data, find the relation of the type y = klQmx
between x and y:
X
i
2
3
4
5
6
7
8
y
0 5
0 8
1 2
1 9
3
4 8
7 5
11 9
169. Differences. Before proceeding to investigate rules
for the choice of the particular type of equation that will repre-
sent the observed values, it is advisable to define and discuss
differences.
Let the observed values be (#„ y»), (i = 0, 1, 2, • • • , n).
The first differences are defined by
(159-1) At/t ss yi+1 — 2/t.
The second differences are given by
A22/t ss A2/l+i — Ay,.
In general, for k > 1, the differences of order k, or the kth
528 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §160
differences, are defined as
(159-2) A*2A s A^-^+i - Ak~lyl.
It should be noted that, if the jth differences are constant, then
all of the differences of order higher than j will be zero.
From (159-1) and (159-2) it follows that
2/i = 2/o + AT/O,
2/2 = 2/i + AS/I = (2/0 + Ai/o) + (A22/0 + AT/O)
= 2/o + 2A2/o + A22/0,
2/3 = 2/2 + ^2/2 = (2/0 + 2A?/o + A2i/o) + (A2?/i + A£/I)
= (2/0 + 2A2/o + A22/0)
+ (A32/0 + A2?/o + A22/0 + A2/0)
= 2/o + 3A2/o + 3A2?/o + A32/o.
These results can be written symbolically as
yl = (1 + A)y0, 2/2 = (1 + A)22/0, 2/3 = (1 + A)3i/o,
in which (1 + A)' acts as an operator on 2/0, with the exponent on
the A indicating the order of the difference. This operator is
analogous to the differential operators discussed in Chap. VII.
By mathematical induction, it is established easily that
(159-3) yk = (1 + A)*y0.
160. Equations That Represent Special Types of Data. There
are certain types of data which suggest the equation that will
represent the relation connecting the observed values of x and y.
Some of the more common types will be discussed in this section.
a. Suppose that a number of pairs of observed values (xt} yl)
have been obtained by experiment. If the xl form an arith-
metical progression and the rth differences of the yl are constant,
then the relation connecting the variables is
y = a0 + dix + a2£2 + • • • + arxr.
For if the rth differences are constant, all differences of order
higher than r are zero, and hence, from (159-3),
(160-1) y* = 2/0
where
(160-2)
§160 EMPIRICAL FORMULAS AND CURVE FITTING 529
is simply the coefficient of ar in the binomial expansion for
(1 + a)k. Moreover, it was assumed that the xt are in arith-
metical progression so that, if x\ — XQ = A#, then XH — x0 = k Ax
and
, _ Xk — Xp
™ - A - *
Ax
Now, the expression (160-2) is a polynomial of degree r in ky and
therefore of degree r in xfc. It follows that, upon substitution of
7 __
K —
Ax
and the collection of like powers of xk in (160-1), this equation
assumes the form
The relation is true for all integral values of &, and therefore
(160-3) y = a0 + aix + a2x2 + • • • + arxr
gives the relation existing between the variables for the given
set of observed values.
In general, a given set of observed values will not possess con-
stant differences of any order, but it may be that the rth differ-
ences are sensibly constant. Then an equation of the type
(160-3) will be a good approximation for the relation between
the variables.
Various modifications of (160-3) can be made. If the values of
x? form an arithmetical progression, whereas the values of the
rth differences of the y? are constant, then the relation connect-
ing the variables is
(160-4) ym = a0 + aixn + a2(zn)2 + • • • + ar(xn)r.
Here m and n can take either positive or negative values. The
derivation of the formula is exactly like that given above if xtn is
replaced by Xl and y? by Ft.
If, in (160-4), m = n = —1 and r = 1, the equation assumes
the form
IT'-o. + a^ or i = «o + or y - -
530 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §160
Curves having this equation are frequently of use in fitting data
to observations measuring flux density against field intensity.
The curve is the hyperbola having the lines
ai , 1
x — ancl y = „
for asymptotes. If the values of l/y are plotted against those
for 1/x, the result is a straight line. A few of these curves are
plotted in Fig. 154.
012 4 6 8
Fio. 154.
Example 1. Consider the following set of observed data:
X
y
&y
A2?/
A3y
A4?/
1
2 105
0 703
2
2 808
0 103
0.806
0 081
3
3 614
0 184
-0 002
0 990
0.079
4
4 004
0 263
-0.001
1 253
0 078
5
5.857
0 341
+0 003
1 594
0 081
6
7 451
0 422
-0.001
2 016
0 080
7
9.467
0.502
2.518
8
11.985
§160
EMPIRICAL FORMULAS AND CURVE FITTING
531
In this case the third differences are sensibly constant and the rela-
tion between x and y is approximately of the form
The question of determining the values of the constants will be con-
sidered in a later section.
b. If the set of pairs of observed values (#t, yl) is such that
the values of #t form an arithme-
tical progression and the corre-
sponding values of ?/t form a
geometrical progression, then the
equation representing the relation
between the variables is
(160-5) y = ka*.
For, taking logarithms of both
sides, the equation becomes
log y = log k + x log a,
which is linear in x and log y.
Hence, if the values xz form an
arithmetical progression, the val-
ues log T/t will do likewise. But
then log y^ — log y^\ = c (for each value of z), so that
- '- = ec = C and
Therefore, the numbers y, form a geometrical progression.
It can be proved that, if the values of the rth differences of
the y% form a geometrical progression when the values of the xr
form an arithmetical progression, then the relation between x
and y is
(160-6) y = a0 + aix + - • • + ar-ixr~l + kax.
If r = 1 in (160-6), the equation becomes
y = «0 + kax.
If a > 1, the values of y increase indefinitely as x is increased.
If a < 1, the curve falls off from its value at x = 0 and approaches
the line y = a0 as an asymptote. Three of these curves are
plotted in Fig. 155.
532 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §160
Example 2. Consider the data given in the following table.
X
1
2
3
4
5
6
7
y
2 157
3
519
4 198
4
539
4 708
4 792
4
835
Ay
1
362
0
679
0
341
0
169 0
084 0
043
Since the first differences have values very nearly equal to the num-
bers that form the geometrical progression whose first term is 1.362
1
2345
Fio 156.
X
and whose ratio is K, the relation between x and y is very neaily of the
form
y = a0 4- kax.
c. The equation
(160-7) y = axn
represents the relation existing between the variables if, when
the x% form a geometrical progression, the yl also form a geo-
metrical progression. For if
x% =
then (160-7) states that
yl = ax?
Hence, the g/» form a geometrical progression whose ratio is
R SB rn.
§160 EMPIRICAL FORMULAS AND CURVE FITTING 533
If the first differences of the 2/< form a geometrical progression
when the Xi form a geometrical progression, then the equation
giving the relation between x and y is
(160-8) * y = k + axn.
For if xt — nrt_i, then (160-8) requires that
A^i = 2A - y^-l = k + ax? — (k + asIU)
and, similarly,
Then
= a#Jli(rn — 1) = arnxJL2(rn — 1)
rn - 1).
and therefore the Ai/t form a geometrical progression whose ratio
is R = *"*.
The curves (160-8) are parabolic if n > 0 and hyperbolic if
n < 0. Three of each type are plotted in Fig. 156.
Example 3. Let the pairs of observed values be
X
0 16
0 4
1 0
2 5
6 25
15 625
y
2
2 210
2 421
2 661
2 929
3 222
The values of xl form a geometrical progression with ratio r = 2.5,
and the values of y% are approximately equal to the terms of the geo-
metrical progression whose first term is 2 and whose ratio is R = 1.1.
Hence, the relation between x and y is very nearly of the type y = axn\
and since R = rn, it follows that 1.1 = (2.5)n or log 1.1 = n log 2.5
and
log 1.1
n =
log 2.5
PROBLEMS
1. A flat surface is exposed to a wind velocity of v miles per hour, and
it is desired to find the relation between v and p, which is the pressure
per square foot on the surface. By experiment the following set of
observed values is obtained. Find the type of formula to fit them.
V
10
15
22 5
33 75
50 625
75 937
p
0 3
0.675
1.519
3.417
7 689
17 300
534 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §161
2. The temperature 6 of a heated body, surrounded by a medium kept
at the constant temperature 0°C., decreases with the time. Find the
kind of formula which expresses the relation between S and t that is
indicated by the following pairs of observed values:
t
0
1
2
3
4
5
6
7
8
0
60 00
51.66
44 46
38 28
32 94
28 32
24 42
21 06
18 06
3. If C represents the number of pounds of coal burned per hour per
square foot of grate and H represents the height of the chimney in
feet, find the type of formula connecting H and (7, using the following
data:
C
19
20
21
22
23
24
25
H
81
90 25
100.00
110 25
121 00
132 25
144 00
161. Constants Determined by Method of Averages. Several
different methods are employed in determining the constants
which appear in the equation that expresses the relation existing
between the variables whose observed values are given. The
method to be described in this section is known as the method
of averages. It is based on certain assumptions concerning the
so-called "residuals" of the observations. Let the pairs of
observed values be (x%, yt), and let y = f(x) be the equation that
represents the relation between x and y for these values. Then,
the expressions
«>» = /(«») - y^
are defined as the residuals of the observations. The method of
averages is based on the assumption that the gum 2^ is zero.
This assumption gives only one condition on the constants
that appear in y = f(x). If there are r of these constants and
if f(x) is linear in them, the further^assumption is made that, if the
residuals are divided into r groups, then 2X = 0 for each group.
This second assumption leads to r equations in the r unknown
constants. It is obvious that different methods of choosing the
groups will lead to different values for the constants. Ordinarily,
the groups are chosen so as to contain approximately the same
number of residuals; and if there are to be k residuals in each
group, the first group contains the first k residuals, the second
group contains the succeeding k residuals, and so on.
§161
EMPIRICAL FORMULAS AND CURVE FITTING
535
A modification of this method is spmetimes used when f(x)
is not linear in its constants, but it will not be discussed here. *
Example. Determine the constants iii the equation that represents
the data given in Example 1, Sec. 160.
It was shown in this example that the equation is of the type
+ OL& 4- &2#2 4- «3#3.
Therefore.
f(x) =
= a0 +
Vl
a0
a0
ao
6ai
#2 + as — 2.105,
4a2 + 8a3 - 2.808,
9a2 + 27a3 - 3.614,
16a2 + 64a3 - 4.604,
25a2 + 125a3 - 5.857,
36a2 + 216a3 - 7.451,
49a2 + 343a3 - 9.467,
64a2 + 512a3 - 11.985.
Let the assumptions be that
vi + v2 = 0, vz + v4 =* 0, t;5
Then the conditions on the constants are
0.
2a0 + 3ai + 5a2 + 9a3 = 4.913,
2a0 + 7ai + 25a2 + 91a3 = 8.218,
2a0 + Hoi + 61a« + 341a3 == 13.308,
2a0 + 15ai + 113a2 + 855a3 = 21.452.
The solution of these equations is
a0 = 1.433, ai = 0.685, a2 = -0.025, a3 = 0.013.
Hence, the equation, as determined by the method of averages, is
y = 1.433 + 0.685z - 0.025z2 + 0.013z8.
PROBLEMS
1. Use the method of averages to find the constants in the equation
y = a0 + aix + a2x2,
which is to represent the given data
X
i
2
3
4
5
6
y
3.13
3 76
6 94
12 62
20 86
31.53
2. Find, by the method of averages, an equation to fit the data
given in Prob. 3 at the end of Sec. 160.
* See SCARBOROUGH, Numerical Analysis.
536 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §162
3. Find the constants in the equation of the type
y = a0 + dix + a&2 + a&*,
which fits the data given in the table
X
i
2
3
4
5
6
7
8
y
3 161
3 073
3 104
3 692
5 513
9 089
15 123
24 091
4. Find, by the method of averages, the constants in the equation
y — a + be* if it is to fit the following data:
X
0 5
1
1 5
2
2 5
3
y
1 630
1 844
2 196
2 778
3 736
5 318
6, Use the method of averages to determine the constants in
y = aex +.b sin x + ex2 so that the equation will represent the data
in the table.
X
0 4
0 6
0 8
1
1 2
1 4
1 6
1 8
2
y
0 258
0 470
0 837
1 392
2 133
3 069
4 225
5 608
7 216
162. Method of Least Squares. This section introduces
another method of determining the constants that appear in the
equation chosen to represent the given data. It is probably
the most useful method and the one most frequently applied.
The two methods already described give different values of the
constants depending upon the judgment of 1^he investigator,
either in reading from a graph or in combining the residuals into
groups. This method has the advantage of giving a unique set
of values to these constants. Moreover, the constants deter-
mined by this method give the "most probable " equation in the
sense that the values of y computed from it are the most probable
values of the observations, it being assumed that the residuals
follow the Gaussian law of error. In short, the principle of least
squares asserts that the best representative curve is that for which
the sum of the squares- of the residuals is a minimum.
Suppose that the given set of observed values (#„ y»), (i =
1, 2, • • • , n), can be represented by the equation
V = /(*)
§162 EMPIRICAL FORMULAS AND CURVE FITTING 537
containing the r undetermined constants ai, a^ • • • , ar. Then,
the n observation equations
y^ = /(**)
are to be solved for these r unknowns. If n = r, there are just
enough conditions to determine the constants; if n < r, there
are not enough conditions and the problem is indeterminate; but,
in general, n > r, and there are more conditions than there are
unknowns. In the general case, the values of the ak which
satisfy any r of these equations will not satisfy the remaining
n — r equations, and the problem is to determine a set of values
of the ak that will give the most probable values of y. Let
v* = & - y*
be th§ residuals, or deviations of the computed values from the
observed values, where y% is the value of y obtained by sub-
stituting x = xl in y = /(#). On the basis of the Gaussian law of
error, the probability of obtaining the observed values yl is
Obviously, P is a maximum when 2 ^? is a minimum.
1 = 1
n
Since S = £ v\ is a function of the r unknowns ai, a2, • • • ,
1 = 1
ar, it follows that necessary conditions for a minimum are
Moreover, each t;» is a function of a^; therefore,
- 2* ! + a* 1
Equations (162-1) are called the normal equations.
If it happens that the r equations (162-1) are linear in the r
unknowns a1; a2; • • * , ar, then these equations can be solved
immediately. This will certainly be the case if f(x) is a poly-
nomial. For let
538 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §162
r r
f(x) = 2 aix'~l so that vl = 2 ^X"1 ~ !/••
Then, dvx/dak = zj'-1, and the normal equations assume the
form, with the aid of (162-2),
(162-3) 2(2 aX-1 - 2/0 tf-1 = 0, (& = 1, 2, • • • , r).
It should be noted that the equation which is obtained by setting
n
k = 1 is S t\ = 0. Reordering the terms in (162-3), so as to
collect the coefficients of the a,, gives
r / n n
Hfi2-4^ Vl V^ r'-4-fc-2\ „ _ ^ r/c_i?. //, _ i o • • • r\
^lU^-'-ty ^V 1 ^\ Xt I Uj — / Xt t/t, \l\j — 1, ^>, j '/•
J=ll=l t=l c
The r linear equations (162-4) can then be solved for the values
of the r unknowns a\, 0,%, - - - , ar.
This procedure may be clarified somewhat by writing out some
of these expressions for a simple specific case. Consider the
data given in the following table. Since the second differences
X
i
2
3
4
y
1 7
1 8
2 3
3 2
of the yt are constant, the equation will have the form
/•/ \ i i n
Then, vl = ai + ^xl + «s^t2 — 2A, and
dvt _ dvt _ dVi __ 2
The normal equations
are
§162 EMPIRICAL FORMULAS AND CURVE FITTING 539
If the coefficients of the a, are collected and the normal equa-
tions put in the form (162-4), one obtains the three equations
4 4
4
a2 + x a3 =
\-i 7 \ = i '
Now,
4
V xt = 1 + 2 + 3 + 4 = 10, ^ x? - 1 + 4 + 9 + 16 = 30,
*»=i
t = 1.7 + 3.6 + 6.9 + 12.8 = 25, etc.
The equations become
4at + 10a2 + 30a3 = 9,
lOai + 30a2 + lOOa* = 25,
30ai + 100a2 + 354a3 = 80.8;
and the solutions are ax = 2, a2 = —0.5, a3 = 0.2.
Even when Eqs. (162-1) are not linear in the unknowns, it may
be possible to solve them easily. However, in most cases it is
convenient to replace the exact residuals by approximate residuals
which are linear in the unknowns. This is accomplished by
expanding y = /(#), treated as a function of ai, a2, • • • , ar, in
Taylor's series in terms of ar — at = Aat, where the at are
approximate values of the at. The values of dt may be obtained
by graphical means or by solving any r of the equations ?/, =
f(xl). The expansion gives
(162-5) y = J(x, ai, • • • , ar) s /(^, di + Aai, • • • , ar + Aar)
/c =* 1
where
Af /)f /)2f ^2f
, etc.
fc«=dfc da j
jfc — d*
540 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §162
Assuming that the dt are chosen so that the Aat are small, the
terms of degree higher than the first can be neglected and (162-5)
becomes
y = f(x, oi, • • • , fir)
The n observation equations are then replaced by the n
approximate equations
(162-6)
If (162-6) is used, the residuals v^ will be linear in the Aa/f, and
hence the resulting conditions, which become c
(162-7)
also will be linear in the Aa*,. Equations (162-7) are called the
normal equations in this case.
In order to illustrate the application of the method of least
squares, two examples will be given. In the first the polynomial
form of f(x) permits the use of (162-4), whereas the second
requires the expansion in Taylor's series.
Example 1. Compute the values of the constants appearing in the
equation of Example 1, Sec. 160.
The equation isy = a0 + a\x + a&2 + azx3, and from the given data
it appears that the normal equations are
8a0 + x>) 01
V
/
( 2} *•') «o + ( 2} ^
\«1 7 xt»l
8 /8\ /8\ /8\ 8
(2) *•') «o + (2) ***) «i + (2) ^5) a2 + (X *»6) «3 = 2) *^..
From the form of the coefficients of the a*, it is seen that it is con-
venient to make a table of the powers of the xt and to form the sums
EMPIRICAL FORMULAS AND CURVE FITTING
541
Sav and 2zt*2/t before attempting to write down the equations in
explicit form.
X*
xS
*t3
z»4
*»5
*»•
1
1
1
1
1
1
2
4
8
16
32
64
3
9
27
81
243
729
4
16
64
256
1,024
4,096
5
25
125
625
3,125
15,625
6
36
216
1,296
7,776
46,656
7
49
343
2,401
16,807
117,649
8
64
512
4,096
32,768
262,144
2*,' 36
204
1,296
8,772
61,776
446,964
xt
y^
#t2A .
xSy,
x\ y\
1
2 105
2.105
2 105
2 105
2
2 808
5 616
11.232
22 464
3
3 614
10 842
32.526
97 578
4
4 604
18 416
73 664
294 656
5
5 857
29 285
146 425
732 125
6
7 451
44 706
268 236
1,609 416
7
9 467
66 269
463 883
3,247 181
8
11 985
95 880
767 040
6,136 320
2o^2/t
47 891
273 119
1,765 111
12,141 845
When the values given in the tables are inserted, the normal equa-
tions become
36a0
204a0
l,296ao +
The solutions are
a0 = 1.426,
36ai
204a!
l,296ai
204a2 + I,296a3 = 47.891,
I,296a2 + 8,772a3 = 273.119,
8,772a2 + 61,776a3 = 1,765.111,
61,776a2 + 446,964a3 = 12,141.845.
= 0.693, a2 = -0.028, a3 = 0.013.
Therefore, the equation, as determined by the method of least squares,
is
y = 1.426 + 0.693z - 0.028z2 + 0.013z3.
It will be observed that these values of the constants are very nearly
the same as those obtained by the method of averages.
542 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §162
Example 2. Compute the constants that appear in the equation
that represents the following data:
t
i
2
3
4
e
51 66
44 46
38 28
32 94
Since the observed values are such that the I* form an arithmetic
progression and the 0t approximately form a geometric progression, the
equation expressing the relation is of the form
6 = ka*.
If the points are plotted on semilogarithmic paper, it is found that
k = 60 and a = 10-° °65 = 0.86, approx. This suggests using k0 = 60
and «o = 0.9 as the first approximations. The first two terms of the
expansion in Taylor's series in terms of Afc = k — 60 and Aa = a — 0.9
are
e = 60(0.9)' + ^Jjb-w AA; + l^ j*-w Aa
= 60(0.9)' -
If the values (£», 0t) are substituted in this equation, four equations
result, namely,
0, « 60(0.9)'» + (0.0)'* AA; + GO^O.g)''-1 Aa, (i = 1, 2, 3, 4).
The problem of obtaining from these four equations the values of Afc
and Aa, which furnish the most probable values of 0t, is precisely the
same as in the case in which the original equation is linear in its con-
stants. The residual equations are
v. = (0.9)'» AA; + 60J,(0.9)'*-1 Aa + 60(0.9)'* - 0t, (i = 1, 2, 3, 4).
Therefore,
4 4
S = £ tv> = V [(0.9)«* Afc + 60M0.9)'*-1 Aa + 60(0.9)'* - 0J2,
and the normal equations
dS
= 0 and
dS
become
=-- 0
60(0.9)<» - (9,]0.
1-1
§162 EMPIRICAL FORMULAS AND CURVE FITTING 543
and
4.
[0.9««AA? + 60e,(0.9)'«-lAa + 60(0.9)'* - 0J60M0.9)''-1 = 0.
When these equations are written in the form
p Afc + q Aa = r,
with all common factors divided out, they are
44
(0.9)2<>A/c + 60 2
4 4
'•-'Aa = T 0t(0.9)<> - 60 V (0.9)««
and
- 60 2^ «,(0.9)«»-1.
As in Example 1, the coefficients are computed most conveniently by
the use of a table.
*»
1
2
3
4
Totals
(0 9)'.
0 9
0 81
0 729
0 6561
(09)2<*
0 81
0 6561
0 531441
0 43046721
2 42800821
*,(0.9)".-*
0 9
1.458
1 77147
1 9131876
6 0426576
t? (09)2<."a
1
3 24
5 9049
8 503056
18 647956
(0t)(09)'.
4C 494
36 0126
27 90612
21.611934
132 024654
(^,)(09)'.-1
51 66
80 028
93.0204
96 05304
320 76144
Substituting the values of the sums from the table gives
2.42800821 AA? + 362.559456 Aa = 132.024654 - 145.6804926
and
6.0426576 A/c + 11 18.87736 Aa = 320.76144 - 362.559456.
Reducing all the numbers to four decimal places gives the following
equations to solve for Afc and Aa:
544 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §163
2.4280 A/c+ 362.5595 Aa = -13.6558,
6.0427 M + 1118.8774 Aa = -41.7980.
The solutions are
A/c = -0.238 and Aa = -0.036.
Hence, the required equation is
0 = 59.762(0.864)'.
PROBLEMS
1. Find the constants in the equation for the data given in Prob. 1,
Sec. 161. Use the method of least squares.
2. Use the method of least squares to determine the values of the
constants in the equation that represents the following data.
X
0 2
0 4
0 6
0 8
1
y
1 25
1 60
2 00
2 50
3 20
3. Apply the method of least squares to the data given in Prob. 1,
Sec. 158.
4. Apply the method of least squares to determine the values of
a and b in Prob. 4, Sec. 161.
163. Method of Moments. Since the method of moments is
one of the most popular methods in use by the statisticians and
economists, a brief discussion of it will be presented. For certain
types of equations, especially those which are linear in their
constants, it provides a simple method of determining the
constants. If the equation has the form
r-l
akxK
this method gives results identical with those obtained by the
method of least squares. In this case the method has a theo-
retical background that justifies its use. When the method is
applied to other types of equations, there is, in general, no such
justification. However, in modified forms it is convenient for
computation and often gives very good results.
Let the set of observed values be (z», t/t), (i = 1, 2, • • • , ri),
and the equation that represents these data be y = /(#). When
the values x = xt are substituted in f(x), there result the corre-
sponding computed values of y, which will be designated by
§164 EMPIRICAL FORMULAS AND CURVE FITTING 545
?/t. The moments of the observed values y% and of the computed
values yt are defined, respectively, by
n n
= 2 x\y^ and M* = 2
1=1 1=1
If f(x) contains r undetermined constants, the method of moments
is based on the assumption that
(163-1) TA = MA, (A = 0, 1, 2, • • • , r - 1).
Since y, is a function of the r undetermined constants, Eqs.
(163-1) give r simultaneous equations in these constants.
The method of moments in this form is most useful when
f(x) is linear in its r constants, so that the r equations (163-1)
can be solved immediately. Various modifications and devices
are used to simplify the computation in case f(x) is not linear in
its constants. These will not be discussed here.*
In the special case in which f(x) is a polynomial, that is,
r-l
f(x) = 2 a^>
j=0
the values of y% are given by
r-l
& = 2 a,x't
and therefore
n r-l
\ = 2 2 x?"a» (* = 0, 1, 2, • • • , r - 1).
1 = 1 j = 0 j=0 i = l
Then Eqs. (163-1) assume the form
2 2 x*+ka< = 2 ^' (fc = 0, 1, 2, • • • , r - 1),
;=0 t = l i = l
which are identical with the normal equations (162-4) obtained
by the method of least squares. Hence, the two methods lead
to identical results for this form of /(#).
164. Harmonic Analysis. The problem of obtaining the
expansion of a periodic function in an infinite trigonometric
* For a discussion, see Frechet and Romann, Representation des lois
empiriques; Rietz, Handbook of Mathematical Statistics.
546 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §164
series was considered in Chap. II. In this section will be given
a short discussion of the problem of fitting a finite trigonometric
sum to a set of observed values (xt, yl) in which the values of y
are periodic.
Let the set of observed values
(X0, ?/o), (Zi, 2/l), ' ' ' , (.T2»-l, 2/2n-l), (Z2n, 2/2»), ' ' '
be such that the values of y start repeating with y^n (that is,
2/2n = ?/o, 2/2»+i = 2/i, etc.). It will be assumed that the xl are
equally spaced, that x0 = 0, and that x*2» = 2ir. [If £0 ^ 0 and
the period is c, instead of 2?r, the variable can be changed by
setting
0t = — (x, — x-0).
c
The discussion would then be carried through for 0t and y% in
place of the xv and ?/, used below.] Under these assumptions
. 2ir ITT
Xi =
The equation
i 1 ^T
2n n
n —
(164-1) y — AO + V Ak cos kx + V Bk sin kx
n — l
V
i ffi
contains the 2n unknown constants
which can be determined so that (164-1) will pass through the
2n given points (xl9 yt) by solving the 2n simultaneous equations
n n— 1
y, = AO + "£ Afc cos fcxt + V ^A, sin fcx,,
ft-i /fe = i
(i = 0, 1, 2, • • • , 2n - 1).
Since xl = zV/n, these equations become
n n-1
/1/>yl rtN . . ^1 . zA:7T . ^%T1 n . zfcir
(164-2) i/ 1= AO + >. Afc cos -- h >, #* sin — ,
(t = 0, 1, 2, • • • , 2n - 1).
The solution of Eqs. (164-2) is much simplified by means of a
scheme somewhat similar to that used in determining the Fourier
§164 EMPIRICAL FORMULAS AND CURVE FITTING 547
coefficients. Multiplying both sides of each equation by its
coefficient of AQ (that is, by unity) and adding the results give
2n-l n 277-1 x n-1 ,271-1
t = (
It can be established that
2n-l
^ cos ^ = 0, (k = 1, 2, • • • , n),
and
277-1
^j sin — = 0, (k = 1, 2, • • • , n — 1).
Therefore,
2n-l
(164-3) 2nA0 = ]£ yt.
Multiplying both sides of each equation by its coefficient of
A]y (j = 1, 2, • • • , n — 1), and adding the results give
^ yl cos - — = ^, ( >, cos — cos -'•— 1 Ak
^J y n ^J \^J n n /
, ^7* /%^ • ikir tjir\
+ >, >, sm — cos — )
^-J \^J n n /
for j = 1, 2, • • • , n - 1. But
cos — - cos — = 0, if k 5* j,
n n ' J)
= n, if k = j; *
and
2n - 1
. ikir ijw A
sm — cos — = 0
>k
n n
for all values of k. Therefore,
(164-4) nA, = "S- y. cos ?A (j - 1, 2; • - • , n - 1).
/&
548 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §164
In order to determine the coefficient of An the procedure is
precisely the same, but
2n-l ^
IKTT ~ .f 7 .
cos — cos ITT — 0, if k 5^ n,
= 2n, if A; = n.
Hence,
2n-J
(164-5) 2nAn = £ y, cos zV.
i = 0
Similarly, by multiplying both sides of each equation of
(164-2) by its coefficient of Rk and adding, it can be established
that
2n-l
(164-6) nB, = ^ »• sin V 0' = 1, 2, - • • , n - 1).
t = 0
Equations (164-3), (164-4), (164-5), and (164-6) give the
solutions for the constants in (164-1). A compact schematic
arrangement is often used to simplify the labor of evaluating
these constants. It will be illustrated in the so-called "6-ordi-
nate" case, that is, when 2n = 6. The method is based on the
equations that determine the constants, together with relations
such as
. TT . (n - l)ir . (n + })TT . (2n - I)TT
sin - = sin • - — = — sin ------ - ~ = — sin — ----- — j
n n n n
TT (n - !)TT (n + l)ir (2/1 - I)TT ,
cos - = ~ cos - - — = — cos -- — = cos - - ~; etc.
n n n n
Six-ordinate Scheme. Here, 2n = 6, the given points are
fo, yt), where xl = nr/3, (i = Ot 1, 2, 3, 4, 5), and Eq. (164-1)
becomes
y = AQ + A\ cos x + A 2 cos 2x + A 3 cos 3# + ^i sin x + Bz sin 2x.
Make the following table of definitions:
2/o y\ 2/2 t>o #1 WQ W\
y\
_ _ _
Sum \VQ v\ Vz Po pi r0 r\
Difference! w0 Wi w2 q\ Si
§164
EMPIRICAL FORMULAS AND CURVE FITTING
549
It can be checked easily that Eqs. (164-3), (164-4), (164-5),
and (164-6), with n = 3, become
6A3 = r0 — si,
3*! =
In particular, suppose that the given points are
X
0
7T
3
27T
3
7T
4r
3
STT
'3
27T
y
1.0
1 4
1 9
1 7
1 5
1 2
1 0
Upon using these values of y in the table of definitions above,
1 0
1 7
1 4
1 5
1 9
1 2
VQ =
2 7
vi = 29
v2 = 31
WQ =
-0 7
MI = -0 1
w;2 = 07
2 7
2 9
-0 7
-0 1
3 1
0 7
PG
,=27
pi = 6 0
r0 = -0 7
ri == 0 6
qi = -0 2
*i == -0.8
Therefore, the equations determining the values of the con-
stants are
3A2 =
2.7 + 6.0 =
0.7 - 0.4 =
2.7 - 3.0 =
0.7 + 0.8 =
8.7
-1.1
-0.3
0.1
(0.6) = 0.3
(-0.2) = -0.1 V3
and
and
and
and
and
and
lo = 1.45,
Li = -0.37,
12 = -0.10,
L3 = 0.02,
?i - 0.17,
h = -0.06.
Hence, the curve of type (164-1) that fits the given data is
y = 1.45 - 0 37 cos x - 0 10 cos 2x -f 0 02 cos 3x + 0.17 sin x - 0.06 sin 2x.
A convenient check upon the computations is furnished by the
relations
and
+ B2 =
(y, - 2/5).
550 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §166
Substituting the values found above in the left-hand members
gives
1.45 - 0.37 - 0.10 + 0.02 = 1.0 and 0.17 - 0.06 -= 0.11,
which check with the values of the right-hand members.
Similar tables can be constructed for 8-ordinates, 12-ordinates,
etc.*
PROBLEMS
1. Use the 6-ordinate scheme to fit a curve of the type (164-1) to the
data in the following table :
X
0
7T
3
27T
3
7T
47T
3
5T
3
2ir
y
0 8
0 6
0 4
0 7
0 9
1 1
• 0 8
2. Make a suitable change of variable, and apply the 6-ordinate
scheme to the data given in the table
X
0
7T
6
*
3
7T
2
27T
~3
5;r
6
7T
y
0 6
0 9
1 3
1 0
0.8
0 5
0 6
165. Interpolation Formulas. When an equation has been
obtained to represent the relation existing between x and y, as
indicated by a given set of observed values (xly yl), this equation
can be used to determine approximately the value of y corre-
sponding to an arbitrary value of x. It would be expected that
the equation would furnish a good approximation to the value
of y corresponding to an x which lies within the range of the
observed values xt. The equation may provide a good approxi-
mation for y even if x is chosen outside this range, but this must
not be assumed.
Frequently, it is desired to obtain an approximation to the y
corresponding to a certain value of x without determining the
relation that connects the variables. Interpolation formulas
have been developed for this purpose and for use in numerical
integration (mechanical quadrature).! The formulas to be
* See CAUSE and SHEARER, A Course in Fourier Analysis and Periodogram
Analysis.
t See Sees. 167 and 168.
§165 EMPIRICAL FORMULAS AND CURVE FITTING 551
discussed here all assume that the desired value for y can be
obtained from the equation
y == a0 + dix + a2x2 + • • • + amxm,
in which the at have been determined so that this equation is
satisfied by m + 1 pairs of the observed values (#,, z/,). These
m + 1 pairs may include the entire set of observed values, or
they may be a subset chosen so that \x — xl is as small as possible.
The first interpolation formula of this discussion assumes that
the set of m + 1 observed values #0, #1, #2, • * * , xm is an
arithmetic progression, that is, that
(165-1) xk = xi^i + d = xo + kd, (k = 1, 2, • • • , m).
Since there are m + 1 pairs of observed values, there is only one
mth difference Awt/, and all differences of order higher than m
are zero. Hence, by (159-3),
k A?/o H --- 2!
*(* - 1) • • • (fc -
But, from (165-1), it follows that
xk - x0
k ~ ~~d~~'
so that the expression for yjc becomes
/Tr.c 0\ . %k — XO * , (%k
(165-2) yk == 7/0 H -- ^ — AT/O H
— So - d) ' ' '
..._ o.
Relation (165-2) is satisfied by every one of the m + 1 pairs
of observed values. Now, assume that the value of the y which
corresponds to an arbitrary x also can be obtained from (165-2).
Then,
,+ n- 0\ i o A , Q — XQ —
(165-3) 2/ = yo + J - j— ^ A2/o + ^ - '-^jj
, (x — x<>)(x — x0 — d) • • • (x — XD — md + d)
' • ' - -
552 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §166
Equation (165-3) represents the mth-degree parabola which
passes through the m + 1 points whose coordinates are (xt, yl).
It assumes a more compact form and is more convenient for
computation purposes when — -. — - is replaced by X. Then,
(165-4) y = i/o + X Ay0 + " A27/0 +
Example. Using the data given in Example 1, Sec. 160, determine
an approximate value for the y corresponding to x = 2.2.
First, let y be determined by using only the two neighboring observed
values (hence, m = 1). Then, x0 = 2, T/O = 2.808, AT/O = 0.806, and
x = M_H? = 0/2. Hence,
y - 2.808 + 0.2(0.806) = 2.969,
which has been reduced to three decimal places because the observed
data are not given more accurately. Obviously, this is simply a
straight-line interpolation by proportional parts.
If the three nearest values are chosen, m = 2, x0 = 1, y0 = 2.105,
A#o = 0.703, A2?/o = 0.103, and X = 2.2 - 1 = 1.2. Then,
y = 2.105 + 1.2(0.703) + ' (0.103) = 2.961,
correct to three decimal places.
If the four nearest values are chosen, m = 3, XQ = 1, y0 — 2.105,
Ay0 = 0.703, A2?/o = 0.103, A37/0 == 0.081, and X = 1.2. Therefore,
y = 2.105 + 1.2(0.703) + (1>2)2(°'2) (0.103)
+ (L2)(°-26)(-°-8) (0.081) = 2.958,
correct to three decimal places.
The value obtained by substituting x — 2.2 in the equation
y = 1.426 + 0.693z - 0.028z2 + 0.013z3,
obtained by the method of least squares (see Example 1, Sec. 162)
is 2.954. It might be expected that a better approximation to this
value could be obtained by choosing m — 4, but investigation shows
that the additional term is too small to affect the third decimal place.
166. Lagrange's Interpolation Formula. The interpolation
formula developed in Sec. 165 applies only when the chosen set
§166 EMPIRICAL FORMULAS AND CURVE FITTING 553
of Xi is an arithmetic progression. If this is not the case, some
other type of formula must be applied.
As in Sec. 165, select the m + 1 pairs of observed values for
which \x — x%\ is as small as possible, and denote them by (xly 7/t),
(i = 0, 1, 2, • • • , m). Let the mth-degree polynomials Pk(x),
(k = 0, 1, 2, • • • , m), be defined by
(166-1) Pk(x) = (*- *o) (*-**)' ' ' (*-*») s JJ (a. _ ^
•C — .XL . -*•
Then, the coefficients ^4& of the equation
m
y=5^(*>
can be»determined so that this equation is satisfied by each of the
m + 1 pairs of observed values (xlf yl). For if x = XL, then
A = ~^i-
* P*(x*)'
since Pk(xi) = 0, if i 5* k. Therefore,
(166-2) y =
is the equation of the mth-degree parabola which passes through
the m + 1 points whose coordinates are (xly yt) . If x is chosen as
any value in the range of the x,, (166-2) determines an approxi-
mate value for the corresponding y.
Equation (166-2) is known as Lagrange's interpolation formula.
Obviously, it can be applied when the x, are in arithmetic progres-
sion, but (165-4) is preferable in that it requires less tedious cal-
culation. Since only one mth-degree parabola can be passed
through m + 1 distinct points, it follows that (165-3), or its
equivalent (165-4), and (166-2) are merely different forms of
the same equation and will furnish the same value for y.
Example. Using the data given in Prob. 1, Sec. 160, apply Lagrange's
formula to find the value of p corresponding to v — 21.
If the two neighboring pairs of observed values are chosen, so that
m — 1,
01 22 ^ 21 1 ^
P = °'675 15^2275 + L519 22-5-=T5 = L350'
correct to three decimal places.
554 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §167
If the three nearest values are chosen, so that m — 2,
« - n.(21 - lfi)(21 -22.5) (21 - 10) (21 - 22.5)
P - "•<* (io _ I5)(io - 22.5) "*" u>0'° (15 - 10)(15 - 22.5)
, , ,1Q (21 - 10)(21 - 15)
+ L519 (22.5 - 10)(22.5 - 15) = L323'
correct to three decimal places.
The value of p obtained from the equation p = 0.003z>2, which repre-
sents the given data, is also 1.323.
PROBLEMS
1. Using the data given in Prob. 2, Sec. 160, find an approximate
value for 6 when t = 2.3. Use m = 1, 2, and 3.
2. Find an approximate value for the y corresponding to x — 2,
using the data given in Example 3, Sec. 160. Use m = I and m = 2.
3. If the observed values are given by the data of Prob. 3, Serc. 160,
find an approximate value of H when C = 21.6. Use m — 1, 2, and 3.
4. Using the data of Prob. 1, Sec. 160, find an approximate value for
p when v = 30. Use m — 1 and m = 2.
167. Numerical Integration.* The definite integral P f(x) dx
is interpreted geometrically as the area under the curve y = f(x)
between the ordinates x = a and x = b. If the function f(x) is
such that its indefinite integral F(x) can be obtained, then from
the fundamental theorem of the integral calculus it follows that
However, if the function f(x) does not possess an indefinite inte-
gral expressible in terms of known functions or if the value of
f(x) is known only for certain isolated values of x, some kind of
approximation formula must be used in order to secure a value
(OT fbj(x) dx.
A formula of numerical integration, or mechanical quadrature,
is one that gives an approximate expression for the value of
fa/(#) dx. The discussion given here is restricted to the case
in which m + 1 pairs of values (xt, yl), or [xl9f(x%)], are given
[either by observation or by computation from y — f(x) if the
form of f(x) is known] and where this set of given values is
represented by (165-3) or (166-2).
The formulas of numerical integration that are most frequently
used are based on the assumption that the xl form an arith-
* For discussion of the accuracy of the formulas given here, see Steffensen,
Interpolation; and Kowalewski, Interpolation und genaherte Quadratur.
§167 EMPIRICAL FORMULAS AND CURVE FITTING 555
metic progression, that is, that xk = XQ + kd. In that Case, all
the m + 1 points (#», yt), (i = 0, 1, 2, • • • , w), lie on the
parabola whose equation is given by (165-3). The area bounded
by the x-axis and this parabola between x = XQ and x = xm
is an approximation to the value of fxx"f(x) dx.
Upon using (165-4) and recalling that
v x —
A = ;
it follows that
(167
-1) ^ y dX = JJ" [2/0 + X Ay, +
°
If m = 1, (167-1) becomes
But
, , , v X — XQ
rf — />» I 'yw/Y Q TJ /i X — .
tH/m ~~~ "H) \^ ifiAJU atH.\ji ^\. — , 7
a
so that
and the formula becomes
(167-2)
If n + 1 pairs of values are given, (167-2) can be applied
successively to the first two pairs, the second and third pairs,
the third and fourth pairs, etc. There results
(167-3) tXnydx= I * y dx + I ** y dx + • • • + I y dx
Jxo Jx* Jxi Jxn-i
_ d d
2 2
d
d ,
— — ^0 -j- 2yi -f* 2t/2 "f~ * * * H" 2?/n_i -)- ?/n)'
2
556 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §167
/
^
^^
Y4
^
Ys
"^
Ye
*
»l »
X
X0 X, X2 X3 X4 X5 X6
FIG. 157.
If m = 2, (167-1) becomes
Formula (167-3) is known as
the trapezoidal rule, for it gives
the value of the sum of the areas
of the n trapezoids whose bases
are the ordinates T/O, yi, y%, ' ' ' ,
*, yn. Figure 157 shows the six
trapezoids in the case of n = 6.
Jo Jo L
X
- J/o) + 3 (2/2 -
1
or
(167-4)
r
ty -^o
Suppose that there are n + 1 pairs of given values, where n is
even. If these n + 1 pairs are divided into the groups of three
pairs with abscissas x^ x2i+i, 2^+2, ( i = 0, 1, • • • , — = — \
then (167-4) can be applied to each group. Hence,
(167-5) fxnydx= f"ydx+ f*'ydx+ - - -
JXO JXQ JXl
= (t/o + 47/i + 2/2) + (2/2 +
y dx
+ 2/4)
d
3
3 [2/0 + yn +
+ 2/3 + • ' • + 2/n-l)
+ 2(7/2 + 7/4 + • • • + 7/n_2)].
Formula (167-5) is known as Simpson's rule with m = 2.
Interpreted geometrically, it gives the value of the sum of the
areas under the second-degree parabolas that have been passed
through the points (x*t 7/2t), (#2l+i, 2/2t+i), and (z2l+2, 7/2l+2), [i = 0,
1, 2, - - - , (n - l)/2].
§167 EMPIRICAL FORMULAS AND CURVE FITTINO 557
If m = 3, (167-1) states that
y dX = y0 + X Aj/o
+ g ^~ A'yo
27 9 .
•)•
9 9
(2/i ~ 2/0) + (2/2 -
o
+ g (2/3 - 82/2 + 82/1 - 2/0)
3
(2/0 + 32/i + 82/2 + 2/3),
or
(167-6) I 3 y dx = ^ (2/0 + 3y, + 82/2 + y,).
J^o o
If n + 1 pairs of values are given and if n is a multiple of 3,
then (167-6) can be applied successively to groups of four pairs
of values to give
CXn ^d
(167-7) ydx = ~[yQ + yn + 3(yi + 2/2 + 2/4 + 2/5 + • • •
J^o O
+ l/n-2 + 2/n-l) + 2(2/3 + 2/6 + ' ' ' + J/n-s)]-
Formula (167-7) is called Simpson's rule with m = 3. It is
not encountered so frequently as (167-3) or (167-5). Other
formulas for numerical integration can be derived by setting
m = 4, 5, • • • in (167-1), but the three given here are sufficient
for ordinary purposes. In most cases, better results are obtained
by securing a large number of observed or computed values, so
that d will be small, and using (167-3) or (167-5).
Example. Using the data given in Example 1, Sec. 160, find an
approximate value for J t y dx.
Using the trapezoidal rule (167-3) gives
y dx = H (2.105 + 5.616 + 7.228 + 9.208 + 11.714
+ 14.902 + 9.467) = 30.120.
558 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §168
Using (167-5) gives
C7 y dx = ^[2.105 + 9.467 + 4(2.808 + 4.604 -f 7.451)
+ 2(3.614 + 5.857)] = 29.989.
Using (167-7) gives
dx = %[2.105 + 9.467 + 3(2.808 + 3.614
+ 5.857 + 7.451) + 2(4.604)] = 29.989.
168. A More General Formula. If numerical integration is
to be used in a problem in which the form of f(x) is known, the
set of values (x^ y) can usually be chosen so that the xl form an
arithmetic progression and one of the formulas of Sec. 167 can
be applied. Even if it is expedient to choose values closer
together for some parts of the range than for other parts, the
formulas of Sec. 167 can be applied successively, with appro-
priate values of d, to those sets of values for which the x^ form an
arithmetic progression. However, if the set of given values was
obtained by observation, it is frequently convenient to use a
formula that does not require that the o?t form an arithmetic
progression.
Suppose that a set of pairs of observed values (x%, yl), (i = 0,
1, 2, • • • , m), is given. The points (xr, 7/t) all lie on the
parabola whose equation is given by (166-2). The area under
this parabola between x = XQ and x = xm is an approximation to
XXm
y dx. The area under the parabola (166-2) is
0
CXa ^ in CXm
(168-1) ydx^y^ =** Pk(x) dx,
in which the expressions for the Pk(x) are given by (166-1).
If m = 1, (168-1) and (166-1) give
(168-2) ydx= -— (x - xO dx
—
CXI
(x - x,) dx
— XQ
Formula (168-2) is identical with (167-2), as would be expected,
but the formula corresponding to (167-3) is
§168 EMPIRICAL FORMULAS AND CURVE FITTING 559
(168-3) f*" y dx = %[(xi - *o)(tfo + J/i) + (x»
If m = 2, (168-1) becomes
(168-4) f" y dx = p^ P (* ~ *iX* - **) dx
Jxt Jro(XQ) Jxo
+ ETT-N G» - so)(z -
"l(^0 J^o
J/L_H!
i(zi) L
3 2
+
,» („_ _,_ ^(,,.2 _ 3.8)
(x0
-*•)]
Formula (168-4) reduces to (167-4) when Xi — XQ = x2 — Xi =
d. The formula that corresponds to (167-5) is too long and
complicated to be of practical importance, and hence it is omitted
here. It is simpler to apply (168-4) successively to groups of
three values and then add the results.
Example. Using the data given in Example 3, Sec. 160, find an
approximate value of | ' y dx.
Using (168-3) determines
y dx = K[0.24(4.21Q) + 0.6(4.631) + 1.5(5.082)
0.16
+ 3.75(5.590)] = 16.187.
560 MATHEMATICS FOR ENGINEERS AND PHYSICISTS §168
Applying (168-4) successively to the first three values and to the
last three values gives
p-2
JoiG
, ^ (0.84)2 [2(1.2 - 0.32 - 1) , 2.210(-0.84)
yax 6 L (-0.24X-0.84) (0.24)(-0.6)
2.421(2 + 0.16 - 1.2)1
J
(0.84) (0.6)
(5.25)2 r 2.421(7.5 - 2 - 6.25) , 2.66(-5.25)
"*" 6 L (-1.5X-5.25) "*" (1.5)(-3.75)
(-1.5X-5.25) " (1.5)(-3.75)
2.929(12.5 + 1 -7.5)1 _ ? Q4
+ - (5.25)(3.75) - J " 17'194'
PROBLEMS
/7
y dx, using the data given
in Example 2, Sec. 160, and applying (167-3). Find the approximate
value if (167-5) is used.
/*50 625
2. Apply (168-3) to determine an approximate value for I p dv,
»/io
using the data given in Prob. 1, Sec. 160.
3. Work the preceding problem by applying (168-4).
4. Apply (167-3) and (167-5) to the data given in Prob. 3, Sec. 160,
in order to determine / H dC.
Jio
5. Find the approximate values of f \/4 + x3 dx obtained by using
x = 0, 1, 2, 3, 4, 5, 6 and applying (167-3) and (167-5).
ANSWERS
CHAPTER I
Pages 14-15
1. (a) convergent; (6) divergent; (c) divergent; (d) convergent;
(e) convergent; (/) divergent; (g) convergent; (h) divergent.
2. (a) convergent; (6) divergent.
4. (a) divergent; (5) convergent; (c) divergent; (d) divergent;
(e) convergent; (/) convergent; (g) divergent; (h) convergent;
(s) convergent; 0) divergent.
Page 22
3. (a) — 1 < x < 1; (6) all finite values; (c) — 1 < x < 1;
(d) x > 1 and x < -1.
4. (a) -^ < x < 4; (6) 0; (c) -I < x < I.
Pages 39-40
1. (a) 1 + x + ~ + ~ + • • ;
,,N X3 . X6 X7 ,
(&) * - 51 + 5! " 71 + ' ' ' J
- , N , x2 , a;4 x«
W X - 2» + 41 " 6» + " ' " '"
#3 x^ x"*
w x ~ 3" + T ~ T + ' ;
(fc)l+x+2i-1j-lir--gr+---.
2. (a) (x - 1) - lA(x - I)2 + K(z ~ I)3 - •
(re - 2)2 . (x - 2)3
(e) 7 -f- 29(a; - 1) + 76(a; - I)2 + H0(* - I)3 -h 90(z - I)4
+ 39(o? - I)5 -f 7(x
4. All finite values of x. 6. x2 < 1.
561
562 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
Pages 45-46
1. 0.984808; 2 - 10-". 3. 0.5446.
2. 9. 6. 2.03617.
8. (a) 0.3103; (6) 0.0201; (c) 0.9461;
w._|V^!-^n + ^n-...;
X " + T " ~ + •••>•(/> °-937; W -0-1026;
j. E! _j_ £! 4. §:L4 4. 2£_5 4_ 37x6 I
v*/ * ~*~ 2! 3! "*" 4! 5! 6! ' ' * '
11. a <> 0.24 radian or 14°.
Pages 53-56
6. 7T/2. 15. 1.05.
6. 214.5 ft.; 25.1 ft. 16. 1 69; 0.881.
14. 2 \/2 E(V2/2, 7T/2) = 3.825.
CHAPTER II
Pages 75-76
16
lirins-gg *
1 — s cos x — 2 ^V 2 _ i cos nx-
n = 2 n
Pages 77-78
. (2n — l)wx -
2
2 A (-l)nH 1 4 <A 1 /0
T i' — n sm n;rx' 2 "" 5? ^ (2ra - I)2 C°S ^ ~" ^*x'
n=l n— 1
18f/7T2 4\ . 7T^ 7T2 . 27TO; /7T2 _ ^\ ^ SlTX
S I I ~1~ 111 ®^^ ~O~ "o" Sin o I I O O T I ^^^ O~
7rttL\l IV d J d \d 6AJ 6
in¥~ • • • ]•
,^36 v (-1)'
cos -3-
CHAPTER III
Page 85
1. (a) 2, -0.75; (6) 1.22,- -0.73; (c) 1.08, -0.55, -0.77; (d) -0.57.
2. 4.49.
ANSWERS 563
Page 91
(a) 1.618, -1, -0.618. (6) 13.968, -6.984 ± 0.29U*.
(c) 3, -1, -1. (d) -1,1,2.
(e) 2 4 v/3 4 V5, 2 + V^a) 4 V^w2, 2 4 AX4a>2 4-
(/) -6, i V5, -f' VS.
Pages 94-96
1. (a) 2, -2, -2; (6) 2, -1, -«; W ±M» «, «'; W) 2, -J*, ±t.
2. (a) (-1, 0), (0, 1), (2, 3); (6) (-3, -2), (-1, 0), (0, 1);
Tc) (-4, -3), (-2, -1), (-1, 0); (d) (-3, -2), (-1, 0), (0, 1),
(2,3)
Page 97
1. 2 924. 6. -0.879, 1 347, 2 532.
2. 1«618, -1, -0.618. 6. -0.418.
3. 2061. 7. 1.226.
4. 1.398.
Pages 101-102
1. 1 226; %. 6. 4.494.
2. 2 310 radians. 6. -0.567.
3. 0.3574, 2.1533. 7. -0.725, 1.221.
4. 0.739.
Pages 106-106
1. 41; -35; 1.
2. (a) (3Ka, 2%3); (&) (1, 0, -1); (c) (5, 4, -3); (d) (1, -^, J$).
Page 114
1. 20; -126; -212.
2. (a) (2, -1, 1); (6) (1, %, -K); (c) (3, -1,2); (d) (1, -1, -2,3).
Pages 121-122
1. (a) (1, —1); (6) inconsistent; (c) inconsistent; (d) (1, 3k — 2, &).
2. (a) (-*/7f5Aj/7,*);(6) (0,0) ; (c) (0,0,0);
(d) (fc/4, 7*/8, *); (e) (fc, 2fc, 0); (/) (0, 0, 0).
CHAPTER IV
Page 126
- * 2 - *
(c) y cos xy 4 1, a cos zy; (d) e* log t/, ex/y;
(e) 2xy 4 A * » a;2.
V 1 — x*
2. (a) 2xy - z\ x* + z,y - 2xz; (b) yz + 4 xz + i
564 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
, N z — zx , x
(c) , — t » sm~x -;
(d) , . x y *
V*2
(x2 + y2 + 22)^' (x2 + y2 + 22)*> (x2 + 2/2 -f z2)^
Pages 129-130
1. 7T/6 cu. ft. 6. $3.46.
2. 11.7ft. 7. 0.112; 0.054.
3. 0.139ft. 8. 53.78; 093.
4. 2250. 9. 0.0037T; 0.3 per cent.
6. 10.85. 10. I.GTTJTT.
Pages 136-136
1. ka*(8 cos 26 + K sin 20). 2. 2r cos 20; -2r2 sin 20.
3. 2r - t;t — 2s; s - r.
7. (a) e^2 (2< sin ^ + I cos ^) ;
(b) 2r(l - 3 tan2 0), -6r2 tan 0 sec2 0.
8. (a) 2z, 2(x + tan x sec2 x);
,.N ^67 . AdV / »dV dV\ dV
(b) cos 0 -r -- h sin 0 - — > r I cos 0 -— — sin 0 - — ]» — —
6x 5?/ \ dy Ox ) dz
Pages 141-143
+ — 9x2 — 4w0 1 — 4t>!/2 e —4uy* — 1 „ y —
12v*(u + v)' 4w2(w + t;)J 4v*(u + w) ' ^ ^TZ
4. 2/gy
'
— xy — uvev — v t —vev+v — x
6. (a) -2, 3, i, -i; (6) ^rqr^ ^r^i' srjr^ snp
-f 4u»' 1 +
13 C^ - sec y , .
v ; 3 sec ?/ tan y + 2zV v y cos 0 - 3-s2 cos 0 - 32
14. 2(* -y):2(* -*):2(y - *).
Pages 146-146
3. J^[3 \/3 + 1 -f «(\/3 + 1)1 or 6.811. 4. 2 \/s2 + 2/2.
ANSWERS 565
Page 149
1. V3/3.
2. (a) 2x + 3y + 2z = 6, ^LzJ = ?^J = * ~ **;
(6) 6z -f 2?/ - 32; = 6, ^-TT- = ^ ^ = g^" ;
c ^ ^Q^ i y°y. \zj& — i 2! / _ ^_^!/ _ \,
a2 62 c2 ' Zo 2/o
Pages 152-163
6. dx/ds = l/\/14, rfi//c?s == 2/V14, d2/d« = 3/VTi. 9. 27°.
Pages 154-155
3. 10^20.
4. /,* cos2 0 +/*» sm 20 -\-fyy sin2 0;
fxxr* sin2 ^ — /xi/r2 sm 20 + /^j/r2 cos2 0 — /xr cos 0 — fyr sin 0.
Pages 157-158
2. e 1 + (x - 1) + (y - 1) + [(x - I)2 + 4 (3 - l)(y - 1)
3. 1 + x + ^ (a:2 - 7/2) + ~ (*• - S^2) + ~
Page 160
1. (a) (3, —26) minimum;
(6) (3, 108) maximum, "(5, 0) minimum;
(c) No maxima 91 minima.
2. x = l/e.
4. (a) cos x — — H and sin re = 0, inflection;
(6) cos x = -^ and sin x = 0, inflection;
(c) sin x =» 0, inflection.
6. (a) # = l/e; (6) x — 1%s, maximum, x = 6^25) inflection,
566 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
Pages 162-163
1. a/3, a/3, a/3 2. 8a6c/3 \/3. 3. a/3, 6/3, r/3
4. A/3P/(2 -v/3 + 3), (-\/3 + l)P/2(2 A/3 + 3), P/(2 V$ + 3).
•^
5. £ = /i = •=— •v/u07r2V', a* = \/5l.
OTT
Pages 170-171
7TC* TTOi .,
TT sin -- cos -~- — 1
1. — 5 1 4. — tan a.
.<&« a:
2. «7r. 6. 2#2.
3' a (l ~~ 10g 2)' 6* 2a7r(a2 ~ 1)~2'
CHAPTER V
Pages 190-191
1. (a/5, a/5). 8. 7ra2/2.
2. 7raV16. 11. 4a2^ - lY
\2 /
4 /7T 2\
3. (a) ?/ du dv; (b) u2v du dv dw. 12. Q a*1 ( - — - ) •
O \4 O /
4. (37ra/16, 0, 0). 13. 8a2
6. 32a3/9 14. x = a cos2 -•
6. (a/4, 6/4, c/4). 15. 7ra4fc/2
7. <T7ra46/2. 16. /a = ^{^abc(a2 -f 62)
Page 195
1. 0. 4. 127ra5/5
CHAPTER VI
Page 199
2. (a) -'%,* (b) -2?/'1'
3. (a) ?^; (6) %; (c) ?£; (rf) ^
4. (a) 0; (6) M; (c) -^5-
Page 202
1. 7ra6. 2. J^. 3. 37ra2/8.
Page 206
I- -Ms- 2. 0. 3. -H2 4. %.
Page 212
1. !%. 3. M.
5. (a) 7T/2; (6) - x/8/4; (c) 1%.
ANSWERS
567
CHAPTER VH
Pages 230-231
1. (y')2 4- 5xy' - y + 5x2 = 0 6. y" - 2y' + 2y =
2. y" 4- y = o. 7. #(y')2 — yy' 4- 1
3. zy'" — y" — xy' 4- y = 0 9. £3y'" — 3z2y" -f
4. zy' 4- (1 — x)y 4- 20"* = 0. - 10. 2xy' - y = 0.
6. (y")2 = [14- (2/')2l3
0.
0.
- 6y - 0.
Pages 264-266
1. 0.417 ft.
2. 0 « 0o +
9. w = vo (1 — e~
11. 0 000667 cal.
Page 268
1. sin"1 y 4- sin"1 x — c.
2. (v - !)/(» + 1) = ce«
3. 2 cos y — sin x cos # 4- x
4. sec # 4- tan y = c.
5. tan"1 y — 2 \/l + x — c.
6. ^6* — ex — \/l — y2 = c
13. rr sin"1 x -f-
14.
16. y — # — log XT/
16. tan"1 y — tan"1
1 . 1
17. -
2/
a; - 1
c.
= c.
7. 1 + y = c(l + x).
8. log [(?/ - l)/y] -f «-* = c.
9. 2 tan"1 e^ + log tanh x/2 * c.
10- - - - - 1(>g 2/ = c.
11. y(2 - log y) = M tan2 x + c.
12. s(l + 4?/2)% = c.
8- - ~ + I +
19.
20.
c(l +x)(l - ty).
7/2 - c(l + ^2).
23. (5 - X)/(A - x) =
2. sin ] (y/x) — log x — c
3. sin (y/x) + log Z = c
4. a;2 — 2zi/ — y2 = c.
5. log y + z3/(3?y3) = c.
11.
Pages 261-262
ft __
7. y =
8. a; =
9. log a
10. x -
13. a:2 + y2 -f
0.
12. y 4- cev/x = 0.
14. « + y + 2 log (2x 4- y - 3) - c.
16. rcy2 = c(x 4- 2y). 19. x + ce*2/<2*<o . o.
17. x3 4- y3 = co;?/. 20. y2 4- 2a;y — x2 = c.
18.
568 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
Pages 264-265
1. sin xy + x2 ™ c. 7. x2 log t/ «=* c.
2. aty + xt/2 4- * - c. 8. (1 - x2)(l - y2) - c.
3. e* -}- x + y — c. 9. ex log y + a;2 = c.
4. x3y — 2/3x = c. 10. Not exact.
5. Not exact. 11. Not exact.
6. sin (y/x) — c. 13. x sin 2?/ = c.
Pages 268-269
1. sin"1 y ± x = c.
2. y = ±^A(x VI - x2 4- sin"1 x) 4- c.
3. x 4- 2/ — tan"1 y = c. 4. y — ee3* = 0, y — ce~x «• 0.
Pages 278-279
1. y « ex. 3. x2 4- wi/2 = c.
2. x2 - yz = c. 9. y = ce*/*.
Pages 280-283
12. p = p0e-kh.
Pages 285-286
cos2 x -F 2(sin x - 1)
6. y — 2 sin x — x cos x H — cos x -\ ---
XX
7. y = 1 + ce1*"-1*. 12. x = ce-2" + | - |.
8. 7 = (E/R)(\ - e-^f/i-). 13. y = tan x - 1 + ce~**n*.
9. y = sin x 4- cex 14. y = (x 4- l)(e* + c).
10. y = c tan x -f e*. 16. y = e3* + ce2*.
11. x = 1 + c<r"2/2.
Page 287
1. ?/ = (48x~2 - 96x~4 - 4) cos x + (IGx"1 - 96x~3) sin x + cx~4.
2. 2/~2 = x + ^ + ce2x. 6. ?T2 = 1 + x2 4- cex*.
3. 2/~6 = %x3 -f ex5. 7. x~2 « y +
4. x = y log ex. 8. t/~J = — 1 4- c
5. y1 = 1 4- log x 4- ex.
Page 291
2. e~aaf f eCa+m)» ^^ 3. e-ax f
ANSWERS 569
Pages 294-295
1. (a) y « ae-'* 4- c2e8*; (6) y - cie3* + c2e2*;
(c) y « (ci + C2x)e~x; (d) y » (r i + c2z)ea! + c8;
(e) y = (ci 4- 02£ 4- c3z2)e~* 4- c4; (/) y » ci cos fcx + c2 sin fcz
+ c3 cosh kx 4- C4 sinh
Pages 298-299
1 v - c e-ax , (^3 - *2) , 2x ^ 2
1. y - cic 4- g + y 27-
2. y - cie-3* + c2e~2j; + ~- 3. y = (ct + c&)e* + x + 2.
4. y = cie-*1 + c2e** + (n)/2P)(x2 - to + 2fc~2), where fc -
5. y = c\ex + c2e~x + Cse2x — x. 6, ?/ = cie"* + c2e*/2 4- 2 sin x.
1. y — (ci + c2x)e:c 4- c$e~2x 4- sin x.
Pages 305-307
1. y = 2 cos Vl%_ViO/(27r); y » 2 cos \Ao* -f VlO sin VIO*.
3. t/ = 10 cos V245«.
4. y = 10e-« (cos A/220^ 4- — ^r sin \722oA; /2 = 400 \/245 dynes.
V \/220 /
5. F - 100 \/2e-600< cos ^500* - |); F = 100e""500^(l -f 500 \/20.
6. 7 = 20 \/5e-50000f(5 sinh 10000 VH>t + \/5 cosh 10000 \/50.
0. 10 -jrj- 4- lOgrz/ = 0; maximum y = -\/3, total drop 2 4- \/3.
Pages 314-315
1. x - %cie' - c2e~« -f %te« - %5e' - ^ - Ke,
y = cie' -f c26~4< 4- K& ~ %t ~ Vis-
3. Cy cloid of radius mE/(eH2).
Pages 317-318
1. No. 2. Yes. 3. No.
Pages 321-322
3. y = -(x2 log a:)/9. 4. y » c^ 4- care 4- z2 4- 1.
Pages 324-325
2. y « CiX-2 4- Ca^-1 4- M log x — %.
3. ?/ = Cix2 4- C2X3. 4. ?/ =» CIXOT 4- cjaj""*"1.
6. y - cix2 4- c2x(5+V2i)/2 4-
6. y - dsU + VSo/a 4- c2x(i-\/30/2 4- *.
3
7. y « do?2 4- c2x — x[(log «)f/2 4- log x].
570 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
Page 329
3. y-cil-a;+-~-h + r,(l -f x + z2 -f
CHAPTER VIII
Pages 356-367
^2! d2 _
^-a;^ = 0;
23 _ 1 dz dz
Page 361
1. (a) ^ = Fi(y -f ox) -h ^a(y - ax);
(b) z ~Fi(y -2x) +F*(y + x);
(e) z = Fi(y - x) -f xF*(y -* x).
2. (a) xy; (c) x*y/2 - a;V3.
Page 372
1. 0.44883; 0.14922; 000004
Pages 375-377
2. 35.5; 41.9.
Pages 385-386
_ 200 -A
* n4i &
n - l)a*»-« r
, „ _ 400 ^ 1
j.2n — I <JMJ^ (^fL —
**• '* £i f^f) 1
n= 1
)(l
4. U = 7, ^.n6~a2*«2'*/o(i
knr), where 1=7-^.
(2n - 1)..
Pages 390-391
V n=l
00
7. / = 0.6 -f- 1.1 2) (~1)ncos
n-l
CHAPTER IX
Pages 398-399
1. 0.5640. 2. (10 V3/3)(i -h j + k).
ANSWERS 571
Pages 403-404
6. 3i + 12j + 4k. 6. 19 V3/3; (\/3/3)(8i + j - 9k).
Page 409
j du _ dudx dudy dudz
' ds ~ ~dxds ~dyds dz ds'
2. (a) jyz + jxz + kxy; (b) i2x -f j2y + k2z;
(c) (x2 + 2/2 + z2)-H(iz + jy + kz);
(d) 2(x2 + 2/2 -f z2)-](iz -f jy + kz).
5. 26 A/2/5 6. 9.
Pages 414-416
1. (a) 3; (b) 2/r; (c) 0
3 i 1 /Mf a. ^Aji . M.A j. i J^ /'Mf 4_ M£ _i_ M_A
da; V dx dy ~*" dz ) ^ J dy \ dx ^~ dy "^ dz )
Pages 420-421
1. (a) 0; (b) 0; (c) 0
Pages 432-433
4. ^ = cij 5. * = x2 —
Page 439
^$ 4. 5.1 ^? _4_ ^?. ?? i ^i^? i J^!_
91 ap + P ~dd + Zl a7 ; 9l a^ + 7 56 >
fcUp ^«i i r
L~^ ""ap J "*" 'pL
(sin gAy) _ dA£\ T 1
~d<t> \ l\_ps
_ _
psmei 08 ~d<t> l_psmO~d<f> p dp
j_ M 1
+ *1p
10. 3p cos 0/r4; 0
CHAPTER X
Pages 443-444
1. (a) 2, 60°; (6) 2 >/2, 45°; (c) 1, 90°.
a (n\ x 4- * -y + ! .. .
y' W x2 -f (y - I)2 +%2-f(2/ -I)2'
*2 + (2/4- I)2 ^ & -f (y -f I)2
572 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
Pages 447-448
4 sin 2x . sinh 2y
cos 2x -h cosh 2y cos 2x -f cosh 2y
6. sin x cosh y -\- ^ cos x sinh y.
7. (a) e-»/«[cos (log \/2) + t sin (log \/2)]j
(b) \/2e~r/* I cos f — j ~ l°g \/2 ) -h * sin ( ~" T ~ l°g "N/2 ) I •
Pages 462-463
3. (a) z*; (b) 1/z; (c) z; (d) log z; (e) cos 2. 7. x3 - 3^2.
Page 461
3. 2 4. Trt. 6. 0. 6. 2wi. 7. 0.
Page 469
2. (a) u =* cos # cosh y, v — —sin # sinh t/;
(b) u = e* cos ?/, v — ex sin y;
(c) u — x — &xy , f :
(d) w = log (x2 4- i/2)J
(e) u = a:/ (a;2 + y2), i
CHAPTER XI
Page 496
l. K; Ks- 2. %0; H; Ko; %8; 1323/46189.
Page 497
1. 33/16660. 4. *%Q.
2. 8H2;H2. 5. %2;«2;%a.
3. M4088-
Pages 499-600
1. 4Mo. 6. n > log 2/(log 6 - log 5).
2. H; «6. 6. ^0; %o.
3. 46413/78125. 7. 16^i5.
4. %. 8. 91854/100000
Page 601
1. 0.775; 0.0000265; $47.50. 2. $11 3. 6.
Page 604
I- («) 12«888; (6) 2%48-
2. (0.65)10 + 10(0.65)9(0.35) + 45(0 65)8(0 35)2 + 120(0 65)7(0.35)3.
3.
4-
Page 608
50; iooC*o(H)100;
ANSWERS 573
Page 512
1. l/\/125T. 2. 200; \/3/(10007r). 3.
Page 616
1. (200)10e-200/10! 2. (20)100e-20/100l
3. 0.136; 0.272; 0.272; 0.181; 0.091; 0.036.
Pages 523-524
4. 0.00896; 0.00850. 5. 0.976; 0.983. 6. First set.
CHAPTER XII
Page 527
1. y =* x/2 + %. 2. y = 2.5s05 8. y - 0.3(10°-2a!).
Pages 633-634
1. p — avn. 2. 0 = fcae.
3. H *= a2C2 -f- aiC -f a0.
Pages 536 6^6
1. y = 4.99 - 3.13z + 1 26z2. 2. H - KC'2 ~ HC' 4- Ji-
3. y = 1.557 + 1 992o: - 0 751a;2 -f 0.100s3.
4. y = 1.3 + 0.2e*. 6. y = 0.3<?« -1.1 sin z + 1.5s2.
Page 544
1. y = 4.99 - 3.13s + 1 26s2. 2. y - 10° 5*.
Page 560
1. ?/ = 0.75 -f- 0.10 cos s — 0 05 cos 3s — 0 29 sin x.
2. ?/ = 0.85 — 0.25 cos 2s — 0.05 cos 4s -f 0.05 cos 6s + 0.26 sin 2s
- 0.03 sin 4s.
Page 654
1. 42 61; 42 50; 42.51. 3. 106.15; 106.09; 106.09.
2. 2.581; 2.627. 4. 2.784; 2.700.
Page 560
1. 25.252; 25.068. 4. 666.25; 666.00.
2. 132.137. 6. 39.30; 38.98.
3. 128.6.
INDEX
Absolute convergence of series, 16,
17, 20, 21
Absolute value of complex number,
441
Addition, of series, 21
of vectors, 393
parallelogram law of, 394
Adiabatic process, 224
Aerodynamics, 133, 431
Algebra, fundamental theorem of, 92
Algebraic theorems, 92-94
Alternating series, 15
am u, 51
Ampere's formula, 52n.
Amplitude of complex number, 441
Amplitude function, 51
Analysis, harmonic, 545
Analytic functions, 451-491
Angle, as a line integral, 195
direction, 146, 398
of lap, 240
of twist, 485
solid, 195
Angular velocity, 61, 191, 236, 404,
424
Applications, of conformal repre-
sentation, 479-491
of line integrals, 217-224
of scalar and vector products,
404-406
Approximate formula, for n!, 509
for probability of most probable
number, 511
in applied mathematics, 55
Approximation, Laplace's or normal,
515
Approximations to binomial law, 512
Arc length, 143
of ellipse, 47
Arc length, of sinusoid, 55
Area, 172
as a double integral, 178
as a line integral, 199-202
element of, 183, 184, 190, 437
positive and negative, 200
surface, 188-196
Argument of complex number, 441
Associative law, for series, 18
for vectors, 394
Asymptotic formula for n\ 509
Asymptotic series, 524
Atmosphere, thickness of, 61
Attraction, law of, 218, 232
motion under, 58, 218
of cone, 196
of cylinder, 196
of sphere, 196, 232
Augmented matrix, 118
Auxiliary equation, 292
Averages, method of, 534
Axes, right- or left-handed, 397
B
Base vectors, 396
Beam, 240-242, 307
Belt on pulley, slipping of, 239
Bending moment, 241
Bernoulh-Euler law, 241, 307
Bernoulli's equation, 286
Bessel functions, 273, 336, 381
expansion in, 339
Bessel's equation, 332, 380
Beta function, 276
Binomial law, 502
approximations to, 512
Binomial series, 40
Biot and Savart, law of, 52
Boundary conditions, 242, 351, 363,
370
Buckling, 299
575
576 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
Cable, flexible, 244
flow of electricity in, 386
supporting horizontal roadway,
242
Cartography, 479
Catenary, 247, 252.
Cauchy-Riemann equations, 221,
450, 455
Cauchy's equation, 322n.
Cauchy's integral formula, 461
Cauchy's integral test, 12
Cauchy's integral theorem, 455
Center of gravity, 177, 182, 183, 187,
190, 191, 196, 522
Change of variables, in derivatives,
154
in integrals, 183-188
Characteristic equation, 292
Charge, distribution of, 487
Charts, distribution, 506
Chemical reaction, 258
Circular functions, 247
Circulation, of a liquid, 475, 477
of a vector, 418, 419
Closed curve, area of, 199-201
direction around, 200
integral around, 201, 203, 206,
216, 421
simple, 200
en u, 51
Coefficients, Fourier, 65
metric, 437
Cofactor, 111, 112
Combinatory analysis, fundamental
principle of, 493
Commutative law, 394, 399, 400
Comparison test for series, 9
Complementary function, 290, 292
Complete elliptic integrals, 48
Complex number, 440
absolute value of, 441
argument of, 441
conjugate of, 444, 488
vector representation of, 440
Complex roots of unity, 87
Complex variable, 440-491
functions of, 444-491
analytic, 451-491
derivative of, 449
integration of, 453
line integral of, 454
Taylor's expansion for, 464
Components of force, 217
Composite function, 134, 137
Condenser, 283, 299, 305, 308, 387
Conditionally convergent series, 16,
17,21
Conditions, Cauchy-Riemann, 221,
450, 455
Dinchlet, 65
for exact differential, 212, 216
Conductivity, 367, 426
Conductor, 486, 489
Conformal mapping, 465, 471
Conformal representation, applica-
tions of, 479-491
Conformal transformation, 467
Conjugate of a complex number,
444, 488
Conjugate functions, 468, 470
Conservation of matter, law of, 429
Conservative field of force, 219, 411
Consistent systems of equations,
117-122
Continuity, equations of, 221, 429,
481
of functions, 23, 28, 124, 448
Contour line, 144
Convergence, absolute, 16, 17, 20,
21, 33
conditional, 16, 17, 21
interval of, 31, 33
of series, 4, 7
tests for, 9, 11, 12, 15, 20, 27, 31,
33
radius of, 31, 33
uniform, 23-30, 33
Cooling, law of, 254
Coordinate lines, 434
Coordinate surfaces, 434
Coordinates, curvilinear, 433-439
cylindrical, 152, 185, 190, 191, 378,
386, 434, 438
INDEX
577
Coordinates, ellipsoidal, 433
parabolic, 439
polar, 183, 184, 276, 279, 386, 438
spherical, 152, 185, 382, 386, 434,
439
cos x, 46, 250
cosh, 247
Cosine, hyperbolic, 247
power series for, 38, 40
Cosine series, 73
Cosines, direction, 146, 147, 151,
188, 194, 398
coth, 249
Cramer's rule, 113
Cross product, 400
Cubic equation, algebraic solution
of,*86
graphical solution of, 83
Curl, 418, 422, 423, 438
Current, 386, 427
Curve, distribution, 504, 516
elastic, 240, 307
map of, 466
Curve fitting, 525-560
Curves, integral, 226, 228, 279
orthogonal, 277, 468
Curvilinear coordinates, 433-439
Cylinder functions (see Bessel func-
tions)
Cylindrical coordinates, 152, 185,
190, 191, 378, 386, 434, 438
D
Dam, gravity, 483
Damping, viscous, 302»
Dead-beat motion, 304
Decomposition of vectors, 396
Definite integrals, 172
change of variable in, 183-188
evaluation of, 172
mean-value theorem for, 210n.
Deflection, 299
Degree of differential equation, 225
Del, V (see Nabla)
Delta amplitude, dn, 51
De Moivre's theorem, 90, 442
Dependence, functional, 2
linear, 116
Dependent events, 495
Derivation of differential equations,
231-247
Derivative, 125
directional, 143, 151, 219
normal, 144, 146, 152
of functions of a complex variable,
449, 452, 463
of hyperbolic functions, 255
of series, 29, 33
partial, 125-143, 153
total, 130-143
Descartes's rule of signs, 94
Determinants, 102-114
cof actors of, 111
expansion of, 106n., Ill
functional or Jacobian, 183
Laplace development of, 111
minors of, 110
of matrix, 115
product of, 110
properties of, 107-112
solution of equations by, 102-114
Wronskian, 317
Deviation, standard, 523
Diagonal term of determinant, 107
Diagram, pv, 223
Differences, 527
Differential, exact, 211, 212, 216,
222, 224, 262, 411, 418, 420
of area, 184, 190
of volume, 185, 187, 190
partial, 128-143
total, 127-143
Differential equations, 225-391
Bernoulli's, 286
BessePs, 332, 380
Cauchy-Riemann, 221, 450, 455
Cauchy's, 322n.
definition of, 225
degree of, 225
derivation of, 231-247
Euler's, 322, 430
exact, 262
first order, 256, 267
Fourier, 425
578 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
Differential equations, general solu-
tion of, 230, 290, 292, 350, 358
homogeneous, 259, 261
homogeneous linear, 290
integral curve of, 226, 228
integrating factor of, 265
Laplace's, 369, 382, 385, 386, 439,
451, 470, 481
Legendre's, 342, 384
linear, 226, 283-349, 357
numerical solution of, 346
of electric circuits, 301, 305, 386
of heat conduction, 367
of membrane, 377
of vibrating spring, 308
of vibrating string, 361
order of, 225
ordinary, 225-349
partial, 225, 350-391
particular integral of, 290, 292,
297, 318, 359
particular solution of, 230
second order, 269, 295
separation of variables in, 257
simultaneous, 312-315
singular solution of, 279
solution in series, 228, 325, 349,
364
solution of, 226
with constant coefficients, 287-
315, 357
with variable coefficients, 284,
315-349
Differential expression, 225
Differential operators, 287-299, 357,
406
Differentiation, of implicit functions,
132-142
of series, 29, 33, 34, 80
partial, 123-171
term by term, 33, 34, 80
under integral sign, 167
Diffusion, 369, 427
DifTusivity, 368w.
Direction angles, 146, 398
Direction components, 146
Direction cosines, 146, 147, 151, 188,
194, 398
Direction ratios, 150, 151
Directional derivative, 143, 151, 219
(See also Gradient)
Dirichlet conditions, 65
Discharge of condenser, 299
Discontinuity, finite, 64
Discriminant of cubic, 89
Distance, element of, 435
Distribution of charge, 487
Distribution charts, 506
Distribution curve, 504, 516
Distributive law, 399, 400
Divergence, of series, 5, 8, 20
/ of a vector, 411, 423, 438
Divergence theorem, 191, 415, 425,
428
dn w, 51
Dot product, 399
Double integrals, 173, 192, 202, 275
Drying of porous solids, 369
Dynamics, laws of, 231
E
e, 42
e™, 250
Effects, superposition of, 129, 223
E(k, <?), 48-51, 54
Elastic curve, 240, 307
Elasticity, 241, 422, 484-486
Electrodynamics, 422, 423n.
Electron, 315
Electrostatic field, 475, 477, 479
Electrostatic force, 487
Electrostatic potential, 487
Electrostatics, 486-491
Element, of arc, 467
of area, 184, 190, 437
of distance, 435
of volume, 185, 187, 190, 437
Elementary functions, 315
expansion of, 35-46, 65-82, 465
Ellipse, area of, 177, 202
center of gravity of, 177
length of arc of, 47
Ellipsoidal coordinates, 433
Elliptic functions, 51
INDEX
579
Elliptic integrals, 47-55
complete, 48
first kind, F(k, v?), 48-55, 238
second kind, E(k, ^), 48-54
third kind, II (n, k, <?), 50
Empirical formulas, 525-560
Entropy, 224
Envelope, 279
Equation, auxiliary, 292
Bernoulli's, 286
Bessel's, 332, 380
characteristic, 292
cubic, 86
Euler, 322
Fourier, 425
mdicial, 334
integral, 347
Laplace's, 195, 369, 382, 385, 386,
439, 451, 470, 481
Legendre's, 342, 384
of continuity, 221, 429, 481
of plane, 147
wave, 432
Equations, Cauchy-Uiemann, 221,
450, 455
consistent, 117-122
dependent, 105
differential, 225-391
Euler's, 430
inconsistent, 105, 117-122
normal, 537, 540
parametric, 143, 149, 150, 199, 215
representing special types of data,
528
simultaneous, 102-122, 139-141
solution of, 83-122
systems of, 102-122
homogeneous linear, 119-122
non-homogeneous linear, 113-
119
Error, Gaussian law of, 520, 536
mean, 516
mean absolute, 522
mean square, 522
of observation, 516
probable, 521
small, 56
Error function, 516
Euler equation, 322
Euler formulas, 78, 251
Euler's equations, 430
Euler's theorem, 136
Evaluation of integrals, by differ-
entiation, 169
in series, 43-46
Even function, 68
Events, dependent, 495
independent, 495
mutually exclusive, 497
Exact differential, 211, 212, 216, 222,
224, 262, 411, 418, 420
Exact differential equation, 262
Expansion, in Bessel functions, 339
in Fourier series, 65-82
in Legcndre polynomials, 346
in Maclaunn's series, 37
in power series, 37-46
in Taylor's series, 37
in trigonometric series, 65
uniqueness of, 38
Expectation, 500
Expected number of successes, 508
Exponential form for trigonometric
functions, 78, 251, 446, 447
Exponential function, expansion for,
42, 446
Extremal values, 164
Extremum, 164
F(k, *>), 48-55
Factor, integrating, 265
Factor theorem, 92
Factorial, n !, approximation for, 509
(See also Gamma functions)
Falling body, 58, 232
Field, 406
conservative, 411
electrostatic, 475, 477, 479
irrotational, 418
Finite discontinuity, 64
Fitting, curve, 525-560
Flexure, 298
Flow, of a liquid, 220, 424, 428, 477,
, 478, 480-484
580 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
Flow, of electricity in a cable, 386
of heat, 256, 368-377, 423, 425
seepage, 483
Fluid motion, 220, 424, 428, 475-
478, 480-484
Flux, 416
Flux density, magnetic, 52
Force, 217, 239, 392, 406, 409
components of, 217
conservative field of, 219
electrostatic, 487
Force function, 411
Forced vibrations, 308, 310
Formula, asymptotic, 509
Cauchy's integral, 461
empirical, 525-560
interpolation, 550
Lagrange's, 552
Poisson, 512
Stirling's, 508
Wallis's, 45
Fourier coefficients, 65
Fourier equation, 425
Fourier series, 63-82
complex form of, 78
differentiation of, 80
expansion in, 65-82
integration of, 80
solution of equations with, 364
Functional dependence, 2
Functional determinant, 183
Functions, 1
analytic, 451
Bessel, 336
Beta, 276
complementary, 290
conjugate, 468, 470
continuous, 23, 448
elementary, 315
expansion of, 35, 65, 155
Gamma, 272-277
holomorphic, 451
homogeneous, 136, 259
hyperbolic, 247-256
of a complex variable, 444-491
of several variables, 123, 160
orthogonal, 81, 339, 345
periodic, 64
Functions, potential, 219, 411
power, 30
real, 2, 123
regular, 451
scalar point, 406
singularities of, 222
stream, 221, 432, 453, *oi
vector point, 406
Fundamental principle, of combina-
tory analysis, 493
of sequences, 6
Fundamental theorem, of algebra, 92
of integral calculus, 172, 457
G
Gamma functions, 272-277
Gauss- Argand diagram, 440
Gaussian law of error, 520, 536
Gauss's theorem, 193
General solution of differential equa-
tion, 230, 290, 292, 350, 358
Geometric series, 9
Gradient, V, 144, 152, 407, 410, 438
Graphical method, of curve fitting,
525
of solution of equations, 83
Gravitational constant, 232
Gravitational law (see Attraction)
Gravitational potential, 219, 408
Gravity, center of, 177, 182, 183,
187, 190, 191, 196, 522
Gravity dam, 483
Green's theorem, for the plane, 202
in space, 191, 418
symmetric form of, 194, 418
H
Harmonic analysis, 545
Harmonic series, 8
Heat conduction, 367
Heat flow, 368-377
equation of, 368, 425
steady, 256, 368, 427
variable, 368, 373, 425
Helix, 151, 152
Holomorphic functions, 451
INDEX
581
Homogeneous equations, differential,
259, 261
linear algebraic, 119-122
linear differential, 290
Homogeneous function, 136, 259
definition of, 136
Euler's theorem on, 136
Hooke's law, 241, 299
Horner's method, 95
Hydrodynamics, 221, 422, 428-433,
480-484
Hyperbola, 247
Hyperbolic functions, 247-256
Hyperbolic paraboloid, 162
I
Imaginary roots, 94
Implicit functions, 132, 137-142
Inclined plane, 280, 282, 306
Incompressible fluid, 424, 430
Inconsistent equations, 105, 117-122
Independence, linear, 116, 317
of path, 208, 216, 452, 455
Independent events, 495
Independent trials, 501
Indicial equation, 334
Infinite series, 1-62
absolute convergence of, 16, 17, 20
conditional convergence of, 16, 17
definition of, 4
expansion in, 35-46, 155-158
of constants, 6-22
of functions, 23-62
of power functions, 30
of trigonometric functions, 63-82
operations on, 21, 29, 33-35
sum of, 4n.
tests for convergence of, 9, 11, 12,
15, 20, 27, 31, 33
theorems on, 17, 21, 27, 28, 29, 31,
33, 34, 36, 38
uniform convergence of, 23-30, 33
Inflection, point of, 159
Initial conditions, 235, 351
Integral calculus, fundamental theo-
rem of, 172, 457
Integral curve, 226, 228, 279
Integral equation, 347
Integral formula, Cauchy's, 461
Integral test for series, 12
Integral theorem, Cauchy's, 455
Integrals, around closed curve, 201,
203, 206, 216, 421
change of variable in, 183-188
definite, 172
double, 173, 192, 202, 275
elliptic, 47-55, 238
evaluation of, by means of series,
43-46
iterated, 175, 180
line, 197-224, 410, 421, 454, 458
mean-value theorem for, 210n.
multiple, 172-196
particular, 290, 292, 297, 318
surface, 188-196, 415, 421
transformation of (see Green's
theorem; Stokes's theorem)
triple, 177, 193
volume, 180, 415
with a parameter, 47, 167
Integrating factor, 265
Integration, by parts, 276
numerical, 554-560
of complex functions, 453
of series, 29, 33, 34, 80
term by term, 33, 34, 80
Interpolation, method of, 101
Interpolation formulas, 550-554
Interval, 4
of convergence, 31, 33
of expansion, 38, 76
Inverse hyperbolic functions, 249,
255
Inversions, 106
Irrotational field, 418, 423, 430
Isolation of roots, 92
Isothermal process, 224
Iterated integrals, 175, 180
Iteration, method of, 297
J»(x\ 336
Jacobian, 141, 183, 190
582 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
K
K9(x), 337
Lagrange's interpolation formula,
552
Lagrange's method of multipliers,
163-167
Lamellar field, 423
Laplace's approximation, 515
Laplace's equation, 195, 369, 382,
385, 386, 439, 451, 470, 481
Law, Bernoulli- Euler, 241
binomial, 502, 512
of attraction, 218
of conservation of matter, 429
of cooling, 254
of dynamics, 231
of error, 520, 536
of gravitation, 232
of small numbers, 512
Least squares, method of, 536
theory of, 521
Legendre polynomials, 344, 384
expansion in, 346
^egendre's equation, 342, 384
Leibnitz's rule (see Differentiation,
under integral sign)
Leibnitz's test (see Test, for alternat-
ing series)
Length, of arc, 143
of ellipse, 47
of sine curve, 55
Level surface, 406
Limit, 2, 124, 454
Line, contour, 144
coordinate, 434
direction cosines of, 146, 147, 151
normal, 144, 146-149
of equal potential, 277
of flow, 475
stream, 277, 432, 467
tangent, 143, 147, 151
vector equation of, 395
Line integrals, 197-224, 410, 421,
454
Line integrals, applications of, 217-
224
around a closed curve, 202, 206,
216, 421
definition of, 197, 454
evaluation of, 202-206, 458
for angle, 195
for area, 201
for work, 217
in space, 215, 410, 421
properties of, 206-217
transformation of, 202, 421
Linear dependence or independence,
116, 317
Linear differential equations, 288-
349, 357
with constant coefficients, 287 357
with variable coefficients, 284,
315-349
Linear differential operator, 287-299
Log z, 446
Logarithmic paper, 526
M
M test, 27
Maclaurin formula, 36
Maclaurin's scries, 37, 249
Magnitude of a vector, 393
Map, geographic, 479
of a curve, 466
Mapping functions, 467
Matrix, 114-122
augmented, 118
determinants of, 115
rank of, 115
Maxima and minima, constrained,
163
for functions of one variable, 158
for functions of several variables,
160
Mean error, 516, 522
Mean- value theorems, 210n.
Measure numbers, 397
Mechanical quadrature, 554
Membrane, vibration of, 377
Mercator's projection, 479
Metric coefficients, 437
INDEX
583
Minima (see Maxima and minima)
Minimax, 162
Modulus, of complex number, 441,
442
of elliptic function, k, 51
Moment, bending, 241
Moment of inertia, 177, 180, 182,
183, 187, 190, 191, 196, 241
Moments, method of, 544
Most probable value, 505
approximation for probability of,
511
Motion, dead-beat, 304
fluid, 220
JAWS of, 231, 234
of a membrane, 377
oscijktory, 304
pendulum, 48, 234
simple harmonic, 233, 301, 314,
380
under gravity, 232
Multiple integrals, 172-196
definition and evaluation of, 173,
179
geometric interpretation of, 177
Multiplication, of complex numbers,
442
of determinants, 110
of series, 21
of vectors, 399
Multiplicity of root, 93, 294
Multiplier, Lagrangian, 165
Multiply connected region, 205, 212,
455
Mutually exclusive events, 497
N
N*bla, or del, V, 194, 195, 407, 414,
422
Newtonian potential, 196
Newton's law, of attraction, 218
of cooling, 254
of dynamics, first law, 231
second law, 231, 272, 363
third law, 231, 234
of gravitation, 232
Newton's method of solution, 97
Normal, to a curve, 144
to a plane, 146, 147
to a surface, 147, 188, 407
Normal approximation, 515
Normal derivative, 144, 146, 152
(See also Gradient)
Normal distribution curve, 516
Normal equations, 537, 540
Normal form, 146
Normal law (see Gaussian law of
error)
Normal line, 144, 146-149
Normal orthogonal functions, 81
Numbers, complex, 440
measure, 397
Numerical integration, 554-560
Numerical solution of differential
equations, 346
O
Odd function, 68
Operator, 528
differential, 287-299, 357, 406
vector (see Curl; Divergence;
Gradient; Nabla)
Order of differential equation, 225
Ordinary differential equations, 225-
• ' 349
(See also Differential equations)
Ordinary discontinuity, 64
Origin of a vector, 393
Orthogonal curves, 277, 468
Orthogonal functions, 81, 339, 345
Orthogonal systems, 434
•Orthogonal trajectories, 277-279
Orthogonal vectors, 398
Oscillation of a spring, 299
Oscillatory motion, 304
Overdamped, 303
p series, 10
Parabola, 244
Parabolic coordinates, 439
Paraboloid, hyperbolic, 162
Parachute, 253, 255
584 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
Parallelogram law of addition, 394
Parameters, 277, 280
integrals containing, 167
variation of, 318
Parametric equations, 143, 149, 150,
199, 215, 247
Partial derivatives, 125-143, 153
Partial differential equation, 350-
391
derivation of, 351
Fourier, 425
integration of, 353
Laplace's, 369, 382, 385, 386, 439
linear, 357
of elastic membrane, 377
of electric circuits, 386
of heat conduction, 367, 425
of vibrating string, 361
Partial differentials, 128-143
Partial differentiation, 123-171
Partial fractions, method of, 297
Partial sum, 4
Particular integral, 290, 292, 297,
318, 359
Particular solution, 230
Path, integrals independent of, 208,
216, 452, 455
Pendulum, simple, 44, 234-238, 306
Periodic function, 64
Picard's method, 347
Plane, equation of, 147
inclined, 280, 282, 306
normal form for, 146
tangent, 146-149
Point, of inflection, 159
singular, 451
Poisson formula, 512
Polar coordinates, 183, 184, 276, 279,
386, 438
Polygon, rectilinear, 478, 485
Polynomials, Legendre, 344, 384
Porous solids, drying of, 369
Potential, electrostatic, 487
gravitational, 219, 408
lines of equal, 277
Newtonian, 196
velocity, 221, 222, 277, 430, 432,
453, 467, 480
Potential function, 219, 411
Power series, 30-62
differentiation of, 33, 34
evaluation of integrals by, 43-46
expansion in, 35-46
functions defined by, 33
integration of, 33, 34
interval of convergence of, 31, 33
operations on, 33-35
theorems on, 31-35
uniform convergence of, 33
uniqueness of expansion in, 38
whose terms are infinite series, 40
Power series solutions of differential
equations, 325-346
Precision constant, 520, 521
Pressure on dam, 484
Primitive, 458
Principal part of increment, 128
Probability, 492-524
Probability curve, 521
Probable error, 521
Probable value, most, 505
probability of, 511
Product, of determinants, 110
scalar, 399
vector, 400
Projection, Mercator's, 479
stereographies, 479
Pulloy, slipping of belt on, 239
pv diagram, 223
Q
Quadrature, mechanical, 554
Quotient, of complex numbers, 444
of power series, 40
R
Radius of convergence, 31, 33
Radius vector, 195
Rank of matrix, 115
Ratio test, 11, 20, 31
Reaction, chemical, 258
Rearrangement of series, 17
Rectilinear polygon, 478, 485
Recursion formula, 273, 328, 331,
334
INDEX
585
Region, multiply connected, 205,
212, 455
of integration, 173
simply connected, 205
Regulafalsi, 101
Regular functions, 451
Remainder in Taylor's series, 36-37
Remainder theorem, 92
Repeated trials, 501
Representation, applications of con-
formal, 479-491
Residuals, 534, 537
Resonance, 310
Riemann surface, 473
Right-handed system of axes, 397
Rod, flow of heat in, 373
vibrations of, 366, 367
Roots, of equations, 83-102
isolation of, 92
theorems on, 92-94
of unity, co, co2, 87
Rot (see Curl)
Rotational field, 418
Rule, Cramer's, 113
Simpson's, 556
trapezoidal, W556
S
Scalar field, 406, 408
Scalar point function, 406, 418
Scalar product, 399
application of, 404
Scalars, 392
Schwartz transformation, 478, 485,
491
Seepage flow, 483
Separation of variables, 257
Sequences, 2
fundamental principle of, 6
limit of, 3
Series, asymptotic, 524
binomial, 40
evaluation of integrals by, 43-46
Fourier, 63-82
infinite, 1-62
integration and differentiation of,
29, 33, 34
Series, of constants, 6-22
of functions, 23-62
power, 30-62
solution of differential equations
by, 228, 325-346
Taylor's and Maclaurin's, 37, 155,
228, 249, 464, 539
tests for convergence of, 9, 11, 12,
15, 20, 27, 31, 33
theorems on, 17, 21, 27, 28, 29, 31,
33, 34, 36, 38
uniform convergence of, 23-30
Shearing stresses, 485
Simple closed curve, 200
Simple harmonic motion, 233, 301,
314, 380
equation of, 234
period of, 234
Simple pendulum, 44, 234-238, 306
Simply connected region, 205
Simpson's rule, 556
Simultaneous differential equations,
312-315
Simultaneous equations, 102-122,
139-141
sin x, 41, 250
sin-1 x, 46
Sine, hyperbolic, 247
length of curve, 55
power series for, 40, 41
Sine series, 73
Singular point, 451
Singular solution, 279
Singularities of function, 222
smh x} 247
Sink (see Source and sink)
Six-ordinate scheme, 548
Slipping of belt on pulley, 239
Small numbers, law of, 512
sn u, 51
Solenoidal field, 423
Solid angle, 195
Solids, drying of porous, 369
Solution, of cubic, 86-91
of differential equations, 226, 228,
325
general, 230, 290, 292, 350, 358
particular, 230
586 MATHEMATICS FOR ENGINEERS AND PHYSICISTS
Solution, of differential equations,
singular, 279
of equations, 8&-122
algebraic, 86, 95
graphical method of, 83
transcendental, 85, 97
of systems of linear algebraic equa-
tions, 102-122
steady-state, 309, 310
Source and sink, 221, 411, 416, 424,
427, 429
Space curves, 149-152
Spherical coordinates, 152, 185, 382,
386, 434, 439
Spring, 299, 308, 313
oscillation of, 299
Standard deviation, 523
Steady heat flow, 256, 368, 369, 427
Steady-state solution, 309, 310
Stereographic projection, 479
Stirling's formula, 508
Stokes's theorem, 421
Stream function, 221, 432, 453, 481
Stream lines, 277, 432, 467
Stresses, shearing, 485
String, vibration of, 361
Sum, of a series, 4n.
of vectors, 393
Superposition of effects, 129, 223
Surface, equation of, 144
level, 406
normal to, 147, 407
Surface integral, 188-196, 415, 421
Surfaces, coordinate, 434
Riemann, 473
Systems of equations, consistent or
inconsistent, 117-122
linear algebraic, 107-122
Taylor's theorem, 36
Tension, 239, 243, 244, 251, 362, 377
Test, Cauchy's integral, 12
comparison, 9
for alternating series, 15
for series, 9, 11, 12, 15, 20, 27, 31,
33
ratio, 11, 20, 31
Weierstrass M, 27
Theory of least squares, 521
Thermodynamics, 222
Torque, 405
Total derivatives, 130-143
Total differential, 127-143
Trajectories, orthogonal, 277-279
Transformation, by analytic func-
tions, 467
conformal, 467
Green's, 191, 202, 418
of element of arc, 467
of integrals, 202
Schwartz, 478, 485, 491
Stokes's, 421
Trapezoidal rule, 556
Trials, repeated and independent,
501
Trigonometric functions, 78, 251,
446, 447
Trigonometric series, 63-82
Triple integrals, 177, 193
U
Undetermined coefficients, 229
Uniform convergence, 23-30
test for, 27
Unit vectors, 394, 397
Unity, roots of, 87
Tangent line, 143, 147, 151
Tangent plane, 14&-149
tanh, 249
Taylor's formula, 35-46, 158
applications of, 41-46
Taylor's series, 37, 464, 539
for functions of two variables,
155-158, 228
Variable, change of, 154, 183-188
complex, 440-491
dependent, 2
independent, 1
Variable heat flow, 368, 373, 425
Variation, of constants (see Varia-
tion, of parameters)
INDEX
587
Variation, of parameters, 318, 323
Vector analysis, 392-439
Vector equation of line, 395
Vector field, 406, 408, 409, 412, 418,
423
Vector point function, 406
Vector product, 400
applications of, 404
Vector relationships, 402
Vectors, 144, 152, 392
addition of, 393
base, 396
curl of, 418
decomposition of, 396
divergence of, 411
magnitude of, 393
multiplication of, 399
origin of, 393
orthogonal, 398
radius, 195
unit, 394, 397
zero, 393
Velocity, 404, 424
angular, 61, 191, 236, 404, 424
critical, 61
of earth's rotation, 61
of escape, 61
terminal, 59, 254
Velocity potential, 221, 222, 277,
430, 432, 453, 467, 480
Vibration, forced, 308, 310
of elastic rod, 366, 367
of membrane, 377
of spring, 308
of string, 361
(See also Simple harmonic mo-
tion)
Viscous damping, 302, 366
Volume, as a triple integral, 180
element of, 185, 187, 190, 437
Volume integral, 180, 415
W
Walhs's formula, 45
Wave equation, 432
Wedge, 473
Weierstrass test, 27
Work, 217, 404
Wronskian, 317
Z
Zero vector, 393
Zonal harmonics (see Legendre poly-
nomials)
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